From fb6131a64e096fc2207bd4ac9cd8cc2fbbf092cb Mon Sep 17 00:00:00 2001
From: yi-ge <a@wyr.me>
Date: Fri, 12 Jun 2020 13:12:29 +0800
Subject: [PATCH] =?UTF-8?q?add:=20=E4=B8=89=E6=95=B0=E4=B9=8B=E5=92=8C?=
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---
 README.md              |  5 +++++
 src/math/3sum.js       | 31 +++++++++++++++++++++++++++++++
 test/math/3sum.test.js |  8 ++++++++
 3 files changed, 44 insertions(+)
 create mode 100644 src/math/3sum.js
 create mode 100644 test/math/3sum.test.js

diff --git a/README.md b/README.md
index 75531d6..b183a22 100644
--- a/README.md
+++ b/README.md
@@ -443,6 +443,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 4. 寻找两个正序数组的中位数 <https://leetcode-cn.com/problems/median-of-two-sorted-arrays/>
   - LintCode 65. 两个排序数组的中位数 <https://www.lintcode.com/problem/median-of-two-sorted-arrays/description>
 
+- [三数之和](src/math/3sum.js)
+
+  - LeetCode 15. 三数之和 <https://leetcode-cn.com/problems/3sum/>
+  - LintCode 57. 三数之和 <https://www.lintcode.com/problem/3sum/description>
+
 ## 堆
 
 - [超级丑数](src/stack/super-ugly-number.js)【未完成】
diff --git a/src/math/3sum.js b/src/math/3sum.js
new file mode 100644
index 0000000..183ae34
--- /dev/null
+++ b/src/math/3sum.js
@@ -0,0 +1,31 @@
+// 一年前做过。复制自：https://leetcode-cn.com/problems/3sum/solution/three-sum-ti-jie-by-wonderful611/
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+export const threeSum = function (nums) {
+  const res = []
+  const length = nums.length
+  nums.sort((a, b) => a - b) // 先排个队，最左边是最弱（小）的，最右边是最强(大)的
+  if (nums[0] <= 0 && nums[length - 1] >= 0) { // 优化1: 整个数组同符号，则无解
+    for (let i = 0; i < length - 2;) {
+      if (nums[i] > 0) break // 优化2: 最左值为正数则一定无解
+      let first = i + 1
+      let last = length - 1
+      do {
+        if (first >= last || nums[i] * nums[last] > 0) break // 两人选相遇，或者三人同符号，则退出
+        const result = nums[i] + nums[first] + nums[last]
+        if (result === 0) { // 如果可以组队
+          res.push([nums[i], nums[first], nums[last]])
+        }
+        if (result <= 0) { // 实力太弱，把菜鸟那边右移一位
+          while (first < last && nums[first] === nums[++first]) { } // 如果相等就跳过
+        } else { // 实力太强，把大神那边右移一位
+          while (first < last && nums[last] === nums[--last]) { }
+        }
+      } while (first < last)
+      while (nums[i] === nums[++i]) { }
+    }
+  }
+  return res
+}
diff --git a/test/math/3sum.test.js b/test/math/3sum.test.js
new file mode 100644
index 0000000..5ac058c
--- /dev/null
+++ b/test/math/3sum.test.js
@@ -0,0 +1,8 @@
+import { threeSum } from '../../src/math/3sum.js'
+
+test('三数之和', () => {
+  expect(threeSum([-1, 0, 1, 2, -1, -4])).toEqual([
+    [-1, -1, 2],
+    [-1, 0, 1]
+  ])
+})