diff --git a/src/main/java/leetcode/editor/cn/ArrayPartitionI.java b/src/main/java/leetcode/editor/cn/ArrayPartitionI.java new file mode 100644 index 0000000..6d24acc --- /dev/null +++ b/src/main/java/leetcode/editor/cn/ArrayPartitionI.java @@ -0,0 +1,64 @@ +//给定长度为 2n 的整数数组 nums ，你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得 +//从 1 到 n 的 min(ai, bi) 总和最大。 +// +// 返回该 最大总和 。 +// +// +// +// 示例 1： +// +// +//输入：nums = [1,4,3,2] +//输出：4 +//解释：所有可能的分法（忽略元素顺序）为： +//1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3 +//2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3 +//3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4 +//所以最大总和为 4 +// +// 示例 2： +// +// +//输入：nums = [6,2,6,5,1,2] +//输出：9 +//解释：最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + +//6 = 9 +// +// +// +// +// 提示： +// +// +// 1 <= n <= 10⁴ +// nums.length == 2 * n +// -10⁴ <= nums[i] <= 10⁴ +// +// Related Topics 贪心 数组 计数排序 排序 👍 260 👎 0 + +package leetcode.editor.cn; + +import java.util.Arrays; + +//561:数组拆分 I +class ArrayPartitionI { + public static void main(String[] args) { + //测试代码 + Solution solution = new ArrayPartitionI().new Solution(); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + public int arrayPairSum(int[] nums) { + Arrays.sort(nums); + int sum = 0; + for (int i = 0; i < nums.length; i = i + 2) { + sum += nums[i]; + } + return sum; + } + } +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/BestTimeToBuyAndSellStockIi.java b/src/main/java/leetcode/editor/cn/BestTimeToBuyAndSellStockIi.java new file mode 100644 index 0000000..8b7c366 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/BestTimeToBuyAndSellStockIi.java @@ -0,0 +1,67 @@ +//给定一个数组 prices ，其中 prices[i] 是一支给定股票第 i 天的价格。 +// +// 设计一个算法来计算你所能获取的最大利润。你可以尽可能地完成更多的交易（多次买卖一支股票）。 +// +// 注意：你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。 +// +// +// +// 示例 1: +// +// +//输入: prices = [7,1,5,3,6,4] +//输出: 7 +//解释: 在第 2 天（股票价格 = 1）的时候买入，在第 3 天（股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。 +// 随后，在第 4 天（股票价格 = 3）的时候买入，在第 5 天（股票价格 = 6）的时候卖出, 这笔交易所能获得利润 = 6-3 = 3 。 +// +// +// 示例 2: +// +// +//输入: prices = [1,2,3,4,5] +//输出: 4 +//解释: 在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。 +// 注意你不能在第 1 天和第 2 天接连购买股票，之后再将它们卖出。因为这样属于同时参与了多笔交易，你必须在再次购买前出售掉之前的股票。 +// +// +// 示例 3: +// +// +//输入: prices = [7,6,4,3,1] +//输出: 0 +//解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。 +// +// +// +// 提示： +// +// +// 1 <= prices.length <= 3 * 10⁴ +// 0 <= prices[i] <= 10⁴ +// +// Related Topics 贪心 数组 动态规划 👍 1353 👎 0 + +package leetcode.editor.cn; + +//122:买卖股票的最佳时机 II +class BestTimeToBuyAndSellStockIi { + public static void main(String[] args) { + //测试代码 + Solution solution = new BestTimeToBuyAndSellStockIi().new Solution(); + System.out.println(solution.maxProfit(new int[]{7, 1, 5, 3, 6, 4})); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + public int maxProfit(int[] prices) { + int result = 0; + for (int i = 1; i < prices.length; i++) { + result += Math.max(0, prices[i] - prices[i - 1]); + } + return result; + } + } +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/FizzBuzz.java b/src/main/java/leetcode/editor/cn/FizzBuzz.java new file mode 100644 index 0000000..dfdc533 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/FizzBuzz.java @@ -0,0 +1,67 @@ +//写一个程序，输出从 1 到 n 数字的字符串表示。 +// +// 1. 如果 n 是3的倍数，输出“Fizz”； +// +// 2. 如果 n 是5的倍数，输出“Buzz”； +// +// 3.如果 n 同时是3和5的倍数，输出 “FizzBuzz”。 +// +// 示例： +// +// n = 15, +// +//返回: +//[ +// "1", +// "2", +// "Fizz", +// "4", +// "Buzz", +// "Fizz", +// "7", +// "8", +// "Fizz", +// "Buzz", +// "11", +// "Fizz", +// "13", +// "14", +// "FizzBuzz" +//] +// +// Related Topics 数学 字符串 模拟 👍 103 👎 0 + +package leetcode.editor.cn; + +import java.util.ArrayList; +import java.util.List; + +//412:Fizz Buzz +class FizzBuzz { + public static void main(String[] args) { + //测试代码 + Solution solution = new FizzBuzz().new Solution(); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + public List fizzBuzz(int n) { + List list = new ArrayList<>(); + for (int i = 1; i <= n; i++) { + if (i % 15 == 0) { + list.add("FizzBuzz"); + } else if (i % 5 == 0) { + list.add("Buzz"); + } else if (i % 3 == 0) { + list.add("Fizz"); + } else { + list.add("" + i); + } + } + return list; + } + } +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/IsSubsequence.java b/src/main/java/leetcode/editor/cn/IsSubsequence.java new file mode 100644 index 0000000..e03fc51 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/IsSubsequence.java @@ -0,0 +1,72 @@ +//给定字符串 s 和 t ，判断 s 是否为 t 的子序列。 +// +// 字符串的一个子序列是原始字符串删除一些（也可以不删除）字符而不改变剩余字符相对位置形成的新字符串。（例如，"ace"是"abcde"的一个子序列，而 +//"aec"不是）。 +// +// 进阶： +// +// 如果有大量输入的 S，称作 S1, S2, ... , Sk 其中 k >= 10亿，你需要依次检查它们是否为 T 的子序列。在这种情况下，你会怎样改变代 +//码？ +// +// 致谢： +// +// 特别感谢 @pbrother 添加此问题并且创建所有测试用例。 +// +// +// +// 示例 1： +// +// +//输入：s = "abc", t = "ahbgdc" +//输出：true +// +// +// 示例 2： +// +// +//输入：s = "axc", t = "ahbgdc" +//输出：false +// +// +// +// +// 提示： +// +// +// 0 <= s.length <= 100 +// 0 <= t.length <= 10^4 +// 两个字符串都只由小写字符组成。 +// +// Related Topics 双指针 字符串 动态规划 👍 504 👎 0 + +package leetcode.editor.cn; + +//392:判断子序列 +class IsSubsequence { + public static void main(String[] args) { + //测试代码 + Solution solution = new IsSubsequence().new Solution(); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + public boolean isSubsequence(String s, String t) { + if (s.length() == 0) { + return true; + } + if (t.length() == 0) { + return false; + } + int index = 0; + for (int i = 0; i < t.length() && index < s.length(); i++) { + if (t.charAt(i) == s.charAt(index)) { + index++; + } + } + return index == s.length(); + } + } +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/MaximumUnitsOnATruck.java b/src/main/java/leetcode/editor/cn/MaximumUnitsOnATruck.java new file mode 100644 index 0000000..ba46e23 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/MaximumUnitsOnATruck.java @@ -0,0 +1,79 @@ +//请你将一些箱子装在 一辆卡车 上。给你一个二维数组 boxTypes ，其中 boxTypes[i] = [numberOfBoxesi, +//numberOfUnitsPerBoxi] ： +// +// +// numberOfBoxesi 是类型 i 的箱子的数量。 +// numberOfUnitsPerBoxi 是类型 i 每个箱子可以装载的单元数量。 +// +// +// 整数 truckSize 表示卡车上可以装载 箱子 的 最大数量 。只要箱子数量不超过 truckSize ，你就可以选择任意箱子装到卡车上。 +// +// 返回卡车可以装载 单元 的 最大 总数。 +// +// +// +// 示例 1： +// +// +//输入：boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4 +//输出：8 +//解释：箱子的情况如下： +//- 1 个第一类的箱子，里面含 3 个单元。 +//- 2 个第二类的箱子，每个里面含 2 个单元。 +//- 3 个第三类的箱子，每个里面含 1 个单元。 +//可以选择第一类和第二类的所有箱子，以及第三类的一个箱子。 +//单元总数 = (1 * 3) + (2 * 2) + (1 * 1) = 8 +// +// 示例 2： +// +// +//输入：boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10 +//输出：91 +// +// +// +// +// 提示： +// +// +// 1 <= boxTypes.length <= 1000 +// 1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000 +// 1 <= truckSize <= 10⁶ +// +// Related Topics 贪心 数组 排序 👍 21 👎 0 + +package leetcode.editor.cn; + +import java.util.Arrays; +import java.util.Comparator; + +//1710:卡车上的最大单元数 +class MaximumUnitsOnATruck { + public static void main(String[] args) { + //测试代码 + Solution solution = new MaximumUnitsOnATruck().new Solution(); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + public int maximumUnits(int[][] boxTypes, int truckSize) { + Arrays.sort(boxTypes, Comparator.comparingInt(o -> o[1])); + int sum = 0; + int index = boxTypes.length - 1; + while (truckSize > 0 && index >= 0) { + if (boxTypes[index][0] < truckSize) { + sum += boxTypes[index][0] * boxTypes[index][1]; + truckSize -= boxTypes[index][0]; + index--; + } else { + sum += truckSize * boxTypes[index][1]; + break; + } + } + return sum; + } + } +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/MergeIntervals.java b/src/main/java/leetcode/editor/cn/MergeIntervals.java new file mode 100644 index 0000000..6c1b1c3 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/MergeIntervals.java @@ -0,0 +1,75 @@ +//以数组 intervals 表示若干个区间的集合，其中单个区间为 intervals[i] = [starti, endi] 。请你合并所有重叠的区间，并返 +//回一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间。 +// +// +// +// 示例 1： +// +// +//输入：intervals = [[1,3],[2,6],[8,10],[15,18]] +//输出：[[1,6],[8,10],[15,18]] +//解释：区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6]. +// +// +// 示例 2： +// +// +//输入：intervals = [[1,4],[4,5]] +//输出：[[1,5]] +//解释：区间 [1,4] 和 [4,5] 可被视为重叠区间。 +// +// +// +// 提示： +// +// +// 1 <= intervals.length <= 10⁴ +// intervals[i].length == 2 +// 0 <= starti <= endi <= 10⁴ +// +// Related Topics 数组 排序 👍 1092 👎 0 + +package leetcode.editor.cn; + +import com.code.leet.entiy.TwoArray; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.List; + +//56:合并区间 +class MergeIntervals{ + public static void main(String[] args) { + //测试代码 + Solution solution = new MergeIntervals().new Solution(); +// TwoArray twoArray = new TwoArray("[[1,4],[2,3]]",true); + TwoArray twoArray = new TwoArray("[[1,3],[2,6],[8,10],[15,18]]",true); + solution.merge(twoArray.getArr()); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) +class Solution { + public int[][] merge(int[][] intervals) { + Arrays.sort(intervals, (o1, o2) -> o1[0]==o2[0]?o1[1]-o2[1]:o1[0]-o2[0]); + List list = new ArrayList<>(); + int[] bef = intervals[0]; + for (int[] arr:intervals) { + if(arr[0]<=bef[1]){ + bef[1] = Math.max(arr[1],bef[1]); + }else{ + list.add(bef); + bef = arr; + } + } + list.add(bef); + int[][] result = new int[list.size()][2]; + for (int i = 0; i < list.size(); i++) { + result[i] = list.get(i); + } + return result; + } +} +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/SearchSuggestionsSystem.java b/src/main/java/leetcode/editor/cn/SearchSuggestionsSystem.java index 7726844..e72df8e 100644 --- a/src/main/java/leetcode/editor/cn/SearchSuggestionsSystem.java +++ b/src/main/java/leetcode/editor/cn/SearchSuggestionsSystem.java @@ -71,81 +71,36 @@ class SearchSuggestionsSystem { //力扣代码 //leetcode submit region begin(Prohibit modification and deletion) class Solution { - List getL; public List> suggestedProducts(String[] products, String searchWord) { Arrays.sort(products); - Map map = new HashMap<>(); - for (String product : products) { - char ch = product.charAt(0); - Node root = map.getOrDefault(ch, new Node(ch)); - Node temp = root; - for (int j = 1; j < product.length(); j++) { - if (temp.child[product.charAt(j) - 'a'] == null) { - temp.child[product.charAt(j) - 'a'] = new Node(product.charAt(j)); - } - temp = temp.child[product.charAt(j) - 'a']; - if (j == product.length() - 1) { - temp.isEnd = true; - temp.str = product; - } - } - map.put(ch, root); - } + int index = 0; + String str = searchWord.substring(0, index + 1); List> result = new ArrayList<>(); - Node node; - if (!map.containsKey(searchWord.charAt(0))) { - for (int i = 0; i < searchWord.length(); i++) { + for (int i = 0; i < products.length; i++) { + if (products[i].startsWith(str)) { + List list = new ArrayList<>(); + for (int j = i; j < Math.min(i + 3, products.length); j++) { + if (products[j].startsWith(str)) { + list.add(products[j]); + } + } + index++; + result.add(new ArrayList<>(list)); + if (index == searchWord.length()) { + break; + } + i--; + str = searchWord.substring(0, index + 1); + } + } + if (index < searchWord.length()) { + int size = result.size(); + for (int i = 0; i < searchWord.length() - size; i++) { result.add(new ArrayList<>()); } - return result; - } - node = map.get(searchWord.charAt(0)); - getL = new ArrayList<>(); - getStr(node); - result.add(new ArrayList<>(getL)); - for (int i = 1; i < searchWord.length(); i++) { - char ch = searchWord.charAt(i); - getL = new ArrayList<>(); - if (node.child[ch - 'a'] == null) { - break; - } - getStr(node.child[ch - 'a']); - node = node.child[ch - 'a']; - result.add(new ArrayList<>(getL)); - } - int size = result.size(); - for (int i = 0; i < searchWord.length() - size; i++) { - result.add(new ArrayList<>()); } return result; } - - public void getStr(Node root) { - if (root.isEnd) { - getL.add(root.str); - } - for (int i = 0; i < 26; i++) { - if (getL.size() == 3) { - break; - } - if (root.child[i] != null) { - getStr(root.child[i]); - } - } - } - - class Node { - char ch; - boolean isEnd; - Node[] child; - String str; - - public Node(char ch) { - this.ch = ch; - child = new Node[26]; - isEnd = false; - } - } } //leetcode submit region end(Prohibit modification and deletion) diff --git a/src/main/java/leetcode/editor/cn/doc/all.json b/src/main/java/leetcode/editor/cn/doc/all.json index 2cc175b..bc5304e 100644 --- a/src/main/java/leetcode/editor/cn/doc/all.json +++ b/src/main/java/leetcode/editor/cn/doc/all.json @@ -1 +1 @@ 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17.25","leaf":true,"level":3,"nodeType":"def","status":"not_started","submitted":0,"title":"单词矩阵","titleSlug":"word-rectangle-lcci"},{"acceptance":0.3418298621904466,"acs":0,"articleLive":2,"columnArticles":49,"formTitle":"[面试题 17.26]稀疏相似度","frequency":17.256487495636375,"frontendQuestionId":"面试题 17.26","leaf":true,"level":3,"nodeType":"def","status":"not_started","submitted":0,"title":"稀疏相似度","titleSlug":"sparse-similarity-lcci"}] \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/doc/content/ArrayPartitionI.md b/src/main/java/leetcode/editor/cn/doc/content/ArrayPartitionI.md new file mode 100644 index 0000000..08791c2 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/doc/content/ArrayPartitionI.md @@ -0,0 +1,35 @@ +

给定长度为 2n 的整数数组 nums ，你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得从 1 到 n 的 min(ai, bi) 总和最大。

+ +

返回该 最大总和 。

+ +

 

+ +

示例 1：

+ +
+输入：nums = [1,4,3,2]
+输出：4
+解释：所有可能的分法（忽略元素顺序）为：
+1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
+2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
+3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
+所以最大总和为 4
+ +

示例 2：

+ +
+输入：nums = [6,2,6,5,1,2]
+输出：9
+解释：最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9
+
+ +

 

+ +

提示：

+ +
    +
  • 1 <= n <= 104
    • +
  • nums.length == 2 * n
    • +
  • -104 <= nums[i] <= 104
    • +
+
Related Topics
  • 贪心
  • 数组
  • 计数排序
  • 排序

  • 👍 260
  • 👎 0
    • \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/doc/content/BestTimeToBuyAndSellStockIi.md b/src/main/java/leetcode/editor/cn/doc/content/BestTimeToBuyAndSellStockIi.md new file mode 100644 index 0000000..313d75a --- /dev/null +++ b/src/main/java/leetcode/editor/cn/doc/content/BestTimeToBuyAndSellStockIi.md @@ -0,0 +1,42 @@ +

    给定一个数组 prices ，其中 prices[i] 是一支给定股票第 i 天的价格。

    + +

    设计一个算法来计算你所能获取的最大利润。你可以尽可能地完成更多的交易（多次买卖一支股票）。

    + +

    注意：你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。

    + +

     

    + +

    示例 1:

    + +
    +输入: prices = [7,1,5,3,6,4]
+输出: 7
+解释: 在第 2 天（股票价格 = 1）的时候买入，在第 3 天（股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。
+     随后，在第 4 天（股票价格 = 3）的时候买入，在第 5 天（股票价格 = 6）的时候卖出, 这笔交易所能获得利润 = 6-3 = 3 。
+
    + +

    示例 2:

    + +
    +输入: prices = [1,2,3,4,5]
+输出: 4
+解释: 在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。
+     注意你不能在第 1 天和第 2 天接连购买股票，之后再将它们卖出。因为这样属于同时参与了多笔交易，你必须在再次购买前出售掉之前的股票。
+
    + +

    示例 3:

    + +
    +输入: prices = [7,6,4,3,1]
+输出: 0
+解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。
    + +

     

    + +

    提示：

    + +
      +
    • 1 <= prices.length <= 3 * 104
      • +
    • 0 <= prices[i] <= 104
      • +
    +
    Related Topics
  • 贪心
  • 数组
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  • 👍 1353
  • 👎 0
    • \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/doc/content/IsSubsequence.md b/src/main/java/leetcode/editor/cn/doc/content/IsSubsequence.md new file mode 100644 index 0000000..776baf2 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/doc/content/IsSubsequence.md @@ -0,0 +1,38 @@ +

    给定字符串 s 和 t ，判断 s 是否为 t 的子序列。

    + +

    字符串的一个子序列是原始字符串删除一些（也可以不删除）字符而不改变剩余字符相对位置形成的新字符串。（例如，"ace"是"abcde"的一个子序列，而"aec"不是）。

    + +

    进阶：

    + +

    如果有大量输入的 S，称作 S1, S2, ... , Sk 其中 k >= 10亿，你需要依次检查它们是否为 T 的子序列。在这种情况下，你会怎样改变代码？

    + +

    致谢：

    + +

    特别感谢 @pbrother 添加此问题并且创建所有测试用例。

    + +

     

    + +

    示例 1：

    + +
    +输入：s = "abc", t = "ahbgdc"
+输出：true
+
    + +

    示例 2：

    + +
    +输入：s = "axc", t = "ahbgdc"
+输出：false
+
    + +

     

    + +

    提示：

    + +
      +
    • 0 <= s.length <= 100
      • +
    • 0 <= t.length <= 10^4
      • +
    • 两个字符串都只由小写字符组成。
      • +
    +
    Related Topics
  • 双指针
  • 字符串
  • 动态规划

  • 👍 504
  • 👎 0
    • \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/doc/content/MaximumUnitsOnATruck.md b/src/main/java/leetcode/editor/cn/doc/content/MaximumUnitsOnATruck.md new file mode 100644 index 0000000..bf27c84 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/doc/content/MaximumUnitsOnATruck.md @@ -0,0 +1,42 @@ +

    请你将一些箱子装在 一辆卡车 上。给你一个二维数组 boxTypes ，其中 boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi] ：

    + +
      +
    • numberOfBoxesi 是类型 i 的箱子的数量。
      • +
    • numberOfUnitsPerBoxi 是类型 i 每个箱子可以装载的单元数量。
      • +
    + +

    整数 truckSize 表示卡车上可以装载 箱子 的 最大数量 。只要箱子数量不超过 truckSize ，你就可以选择任意箱子装到卡车上。

    + +

    返回卡车可以装载 单元 的 最大 总数。

    + +

     

    + +

    示例 1：

    + +
    +输入：boxTypes = [[1,3],[2,2],[3,1]], truckSize = 4
+输出：8
+解释：箱子的情况如下：
+- 1 个第一类的箱子，里面含 3 个单元。
+- 2 个第二类的箱子，每个里面含 2 个单元。
+- 3 个第三类的箱子，每个里面含 1 个单元。
+可以选择第一类和第二类的所有箱子，以及第三类的一个箱子。
+单元总数 = (1 * 3) + (2 * 2) + (1 * 1) = 8
    + +

    示例 2：

    + +
    +输入：boxTypes = [[5,10],[2,5],[4,7],[3,9]], truckSize = 10
+输出：91
+
    + +

     

    + +

    提示：

    + +
      +
    • 1 <= boxTypes.length <= 1000
      • +
    • 1 <= numberOfBoxesi, numberOfUnitsPerBoxi <= 1000
      • +
    • 1 <= truckSize <= 106
      • +
    +
    Related Topics
  • 贪心
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  • 排序

  • 👍 21
  • 👎 0
    • \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/doc/content/MergeIntervals.md b/src/main/java/leetcode/editor/cn/doc/content/MergeIntervals.md new file mode 100644 index 0000000..74cb540 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/doc/content/MergeIntervals.md @@ -0,0 +1,29 @@ +

    以数组 intervals 表示若干个区间的集合，其中单个区间为 intervals[i] = [starti, endi] 。请你合并所有重叠的区间，并返回一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间。

    + +

     

    + +

    示例 1：

    + +
    +输入：intervals = [[1,3],[2,6],[8,10],[15,18]]
+输出：[[1,6],[8,10],[15,18]]
+解释：区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].
+
    + +

    示例 2：

    + +
    +输入：intervals = [[1,4],[4,5]]
+输出：[[1,5]]
+解释：区间 [1,4] 和 [4,5] 可被视为重叠区间。
    + +

     

    + +

    提示：

    + +
      +
    • 1 <= intervals.length <= 104
      • +
    • intervals[i].length == 2
      • +
    • 0 <= starti <= endi <= 104
      • +
    +
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  • 👍 1092
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    • \ No newline at end of file