259:较小的三数之和

src/main/java/leetcode/editor/cn/ThreeSumSmaller.java

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+//给定一个长度为 n 的整数数组和一个目标值 target ，寻找能够使条件 nums[i] + nums[j] + nums[k] < target 成立的
+//三元组 i, j, k 个数（0 <= i < j < k < n）。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入: nums = [-2,0,1,3], target = 2
+//输出: 2 
+//解释: 因为一共有两个三元组满足累加和小于 2:
+//     [-2,0,1]
+//     [-2,0,3]
+// 
+//
+// 示例 2： 
+//
+// 
+//输入: nums = [], target = 0
+//输出: 0 
+//
+// 示例 3： 
+//
+// 
+//输入: nums = [0], target = 0
+//输出: 0 
+//
+// 
+//
+// 提示: 
+//
+// 
+// n == nums.length 
+// 0 <= n <= 3500 
+// -100 <= nums[i] <= 100 
+// -100 <= target <= 100 
+// 
+// Related Topics 数组 双指针 二分查找 排序 👍 106 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.Arrays;
+
+//259:较小的三数之和
+public class ThreeSumSmaller {
+    public static void main(String[] args) {
+        Solution solution = new ThreeSumSmaller().new Solution();
+        // TO TEST
+        solution.threeSumSmaller(new int[]{-1, 1, -1, -1}, -1);
+    }
+
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int threeSumSmaller(int[] nums, int target) {
+            if (nums == null || nums.length == 0) {
+                return 0;
+            }
+            Arrays.sort(nums);
+            int sum;
+            int count = 0;
+            for (int i = 0; i < nums.length; i++) {
+                int left = i + 1;
+                int right = nums.length - 1;
+                while (left < right) {
+                    sum = nums[i] + nums[left] + nums[right];
+                    if (sum < target) {
+                        count += (right - left);
+                        left++;
+                    } else {
+                        right--;
+                    }
+                }
+            }
+            return count;
+        }
+    }
+
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/doc/content/ThreeSumSmaller.md

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+<p>给定一个长度为 <code>n</code> 的整数数组和一个目标值 <code>target</code>&nbsp;，寻找能够使条件&nbsp;<code>nums[i] + nums[j] + nums[k] &lt; target</code>&nbsp;成立的三元组&nbsp; <code>i, j, k</code>&nbsp;个数（<code>0 &lt;= i &lt; j &lt; k &lt; n</code>）。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入: </strong><em>nums</em> = <code>[-2,0,1,3]</code>, <em>target</em> = 2
+<strong>输出: </strong>2 
+<strong>解释: </strong>因为一共有两个三元组满足累加和小于 2:
+&nbsp;    [-2,0,1]
+     [-2,0,3]
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入: </strong><em>nums</em> = <code>[]</code>, <em>target</em> = 0
+<strong>输出: </strong>0 </pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入: </strong><em>nums</em> = <code>[0]</code>, <em>target</em> = 0
+<strong>输出: </strong>0 </pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>n == nums.length</code></li>
+	<li><code>0 &lt;= n &lt;= 3500</code></li>
+	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
+	<li><code>-100 &lt;= target &lt;= 100</code></li>
+</ul>
+<div><div>Related Topics</div><div><li>数组</li><li>双指针</li><li>二分查找</li><li>排序</li></div></div><br><div><li>👍 106</li><li>👎 0</li></div>