diff --git a/src/main/java/leetcode/editor/cn/RandomPickWithWeight.java b/src/main/java/leetcode/editor/cn/RandomPickWithWeight.java new file mode 100644 index 0000000..64033a5 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/RandomPickWithWeight.java @@ -0,0 +1,108 @@ +//给定一个正整数数组 w ,其中 w[i] 代表下标 i 的权重(下标从 0 开始),请写一个函数 pickIndex ,它可以随机地获取下标 i,选取下标 +//i 的概率与 w[i] 成正比。 +// +// +// +// +// 例如,对于 w = [1, 3],挑选下标 0 的概率为 1 / (1 + 3) = 0.25 (即,25%),而选取下标 1 的概率为 3 / (1 + +// 3) = 0.75(即,75%)。 +// +// 也就是说,选取下标 i 的概率为 w[i] / sum(w) 。 +// +// +// +// 示例 1: +// +// 输入: +//["Solution","pickIndex"] +//[[[1]],[]] +//输出: +//[null,0] +//解释: +//Solution solution = new Solution([1]); +//solution.pickIndex(); // 返回 0,因为数组中只有一个元素,所以唯一的选择是返回下标 0。 +// +// 示例 2: +// +// 输入: +//["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"] +//[[[1,3]],[],[],[],[],[]] +//输出: +//[null,1,1,1,1,0] +//解释: +//Solution solution = new Solution([1, 3]); +//solution.pickIndex(); // 返回 1,返回下标 1,返回该下标概率为 3/4 。 +//solution.pickIndex(); // 返回 1 +//solution.pickIndex(); // 返回 1 +//solution.pickIndex(); // 返回 1 +//solution.pickIndex(); // 返回 0,返回下标 0,返回该下标概率为 1/4 。 +// +//由于这是一个随机问题,允许多个答案,因此下列输出都可以被认为是正确的: +//[null,1,1,1,1,0] +//[null,1,1,1,1,1] +//[null,1,1,1,0,0] +//[null,1,1,1,0,1] +//[null,1,0,1,0,0] +//...... +//诸若此类。 +// +// +// +// +// 提示: +// +// +// 1 <= w.length <= 10000 +// 1 <= w[i] <= 10^5 +// pickIndex 将被调用不超过 10000 次 +// +// Related Topics 数学 二分查找 前缀和 随机化 👍 143 👎 0 + +package leetcode.editor.cn; + +//528:按权重随机选择 +class RandomPickWithWeight { + public static void main(String[] args) { + //测试代码 +// Solution solution = new RandomPickWithWeight().new Solution(); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + private int[] arr; + private int[] weight; + + public Solution(int[] w) { + arr = w; + weight = new int[w.length]; + weight[0] = arr[0]; + for (int i = 1; i < arr.length; i++) { + weight[i] = weight[i - 1] + arr[i]; + } + } + + public int pickIndex() { + int start = 0; + int end = weight.length - 1; + int max = (int) (Math.random() * weight[end]) + 1; + while (start < end) { + int mid = (end - start) / 2 + start; + if (weight[mid] < max) { + start = mid + 1; + } else { + end = mid; + } + } + return start; + } + } + +/** + * Your Solution object will be instantiated and called as such: + * Solution obj = new Solution(w); + * int param_1 = obj.pickIndex(); + */ +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/doc/all.json b/src/main/java/leetcode/editor/cn/doc/all.json index 663c99d..3ffec4d 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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b/src/main/java/leetcode/editor/cn/doc/content/RandomPickWithWeight.md @@ -0,0 +1,57 @@ +

给定一个正整数数组 w ,其中 w[i] 代表下标 i 的权重(下标从 0 开始),请写一个函数 pickIndex ,它可以随机地获取下标 i,选取下标 i 的概率与 w[i] 成正比。

+ +
    +
+ +

例如,对于 w = [1, 3],挑选下标 0 的概率为 1 / (1 + 3) = 0.25 (即,25%),而选取下标 1 的概率为 3 / (1 + 3) = 0.75(即,75%)。

+ +

也就是说,选取下标 i 的概率为 w[i] / sum(w)

+ +

 

+ +

示例 1:

+ +
输入:
+["Solution","pickIndex"]
+[[[1]],[]]
+输出:
+[null,0]
+解释:
+Solution solution = new Solution([1]);
+solution.pickIndex(); // 返回 0,因为数组中只有一个元素,所以唯一的选择是返回下标 0。
+ +

示例 2:

+ +
输入:
+["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
+[[[1,3]],[],[],[],[],[]]
+输出:
+[null,1,1,1,1,0]
+解释:
+Solution solution = new Solution([1, 3]);
+solution.pickIndex(); // 返回 1,返回下标 1,返回该下标概率为 3/4 。
+solution.pickIndex(); // 返回 1
+solution.pickIndex(); // 返回 1
+solution.pickIndex(); // 返回 1
+solution.pickIndex(); // 返回 0,返回下标 0,返回该下标概率为 1/4 。
+
+由于这是一个随机问题,允许多个答案,因此下列输出都可以被认为是正确的:
+[null,1,1,1,1,0]
+[null,1,1,1,1,1]
+[null,1,1,1,0,0]
+[null,1,1,1,0,1]
+[null,1,0,1,0,0]
+......
+诸若此类。
+
+ +

 

+ +

提示:

+ + +
Related Topics
  • 数学
  • 二分查找
  • 前缀和
  • 随机化

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