From dddbb95ea718c7dfd327b270b4686a745d1e848d Mon Sep 17 00:00:00 2001
From: Yuhao Ju <juguaguaji@gmail.com>
Date: Thu, 24 Nov 2022 22:05:59 +0800
Subject: [PATCH] =?UTF-8?q?update=200015.=E4=B8=89=E6=95=B0=E4=B9=8B?=
 =?UTF-8?q?=E5=92=8C=EF=BC=9A=E5=AE=8C=E5=96=84=E6=B3=A8=E9=87=8A=EF=BC=8C?=
 =?UTF-8?q?=E4=BC=98=E5=8C=96go=E4=BB=A3=E7=A0=81=E9=A3=8E=E6=A0=BC?=
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---
 problems/0015.三数之和.md | 56 +++++++++++++++++++++--------------
 1 file changed, 33 insertions(+), 23 deletions(-)

diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md
index 5ee57127..6675e408 100644
--- a/problems/0015.三数之和.md
+++ b/problems/0015.三数之和.md
@@ -253,13 +253,15 @@ class Solution {
     public List<List<Integer>> threeSum(int[] nums) {
         List<List<Integer>> result = new ArrayList<>();
         Arrays.sort(nums);
-
+	// 找出a + b + c = 0
+        // a = nums[i], b = nums[left], c = nums[right]
         for (int i = 0; i < nums.length; i++) {
-            if (nums[i] > 0) {
+	    // 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
+            if (nums[i] > 0) { 
                 return result;
             }
 
-            if (i > 0 && nums[i] == nums[i - 1]) {
+            if (i > 0 && nums[i] == nums[i - 1]) {  // 去重a
                 continue;
             }
 
@@ -273,7 +275,7 @@ class Solution {
                     left++;
                 } else {
                     result.add(Arrays.asList(nums[i], nums[left], nums[right]));
-
+		    // 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
                     while (right > left && nums[right] == nums[right - 1]) right--;
                     while (right > left && nums[left] == nums[left + 1]) left++;
                     
@@ -294,12 +296,15 @@ class Solution:
         ans = []
         n = len(nums)
         nums.sort()
+	# 找出a + b + c = 0
+        # a = nums[i], b = nums[left], c = nums[right]
         for i in range(n):
             left = i + 1
             right = n - 1
-            if nums[i] > 0:
+	    # 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
+            if nums[i] > 0: 
                 break
-            if i >= 1 and nums[i] == nums[i - 1]:
+            if i >= 1 and nums[i] == nums[i - 1]: # 去重a
                 continue
             while left < right:
                 total = nums[i] + nums[left] + nums[right]
@@ -309,13 +314,14 @@ class Solution:
                     left += 1
                 else:
                     ans.append([nums[i], nums[left], nums[right]])
+		    # 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
                     while left != right and nums[left] == nums[left + 1]: left += 1
                     while left != right and nums[right] == nums[right - 1]: right -= 1
                     left += 1
                     right -= 1
         return ans
 ```
-Python (v2):
+Python (v3):
 
 ```python
 class Solution:
@@ -344,32 +350,36 @@ class Solution:
 Go：
 
 ```Go
-func threeSum(nums []int)[][]int{
+func threeSum(nums []int) [][]int {
 	sort.Ints(nums)
-	res:=[][]int{}
-	
-	for i:=0;i<len(nums)-2;i++{
-		n1:=nums[i]
-		if n1>0{
+	res := [][]int{}
+	// 找出a + b + c = 0
+	// a = nums[i], b = nums[left], c = nums[right]
+	for i := 0; i < len(nums)-2; i++ {
+		// 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
+		n1 := nums[i]
+		if n1 > 0 {
 			break
 		}
-		if i>0&&n1==nums[i-1]{
+		// 去重a
+		if i > 0 && n1 == nums[i-1] {
 			continue
 		}
-		l,r:=i+1,len(nums)-1
-		for l<r{
-			n2,n3:=nums[l],nums[r]
-			if n1+n2+n3==0{
-				res=append(res,[]int{n1,n2,n3})
-				for l<r&&nums[l]==n2{
+		l, r := i+1, len(nums)-1
+		for l < r {
+			n2, n3 := nums[l], nums[r]
+			if n1+n2+n3 == 0 {
+				res = append(res, []int{n1, n2, n3})
+				// 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
+				for l < r && nums[l] == n2 {
 					l++
 				}
-				for l<r&&nums[r]==n3{
+				for l < r && nums[r] == n3 {
 					r--
 				}
-			}else if n1+n2+n3<0{
+			} else if n1+n2+n3 < 0 {
 				l++
-			}else {
+			} else {
 				r--
 			}
 		}