From c430004a0b8b91419e43bb648b0b2c3b5a8b5db2 Mon Sep 17 00:00:00 2001
From: X-shuffle <53906918+X-shuffle@users.noreply.github.com>
Date: Sun, 20 Jun 2021 14:47:48 +0800
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添加 0131.分割回文串 go版本
---
 problems/0131.分割回文串.md | 44 ++++++++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)

diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md
index 9d23fd13..43453409 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -312,6 +312,50 @@ class Solution:
 ```
 
 Go：
+> 注意切片（go切片是披着值类型外衣的引用类型）
+
+```go
+func partition(s string) [][]string {
+    var tmpString []string//切割字符串集合
+    var res [][]string//结果集合
+    backTracking(s,tmpString,0,&res)
+    return res
+}
+func backTracking(s string,tmpString []string,startIndex int,res *[][]string){
+    if startIndex==len(s){//到达字符串末尾了
+        //进行一次切片拷贝，怕之后的操作影响tmpString切片内的值
+        t := make([]string, len(tmpString))
+		copy(t, tmpString)
+        *res=append(*res,t)
+    }
+    for i:=startIndex;i<len(s);i++{
+        //处理（首先通过startIndex和i判断切割的区间，进而判断该区间的字符串是否为回文，若为回文，则加入到tmpString，否则继续后移，找到回文区间）（这里为一层处理）
+        if isPartition(s,startIndex,i){
+            tmpString=append(tmpString,s[startIndex:i+1])
+        }else{
+            continue
+        }
+        //递归
+        backTracking(s,tmpString,i+1,res)
+        //回溯
+        tmpString=tmpString[:len(tmpString)-1]
+    }
+}
+//判断是否为回文
+func isPartition(s string,startIndex,end int)bool{
+    left:=startIndex
+    right:=end
+    for ;left<right;{
+        if s[left]!=s[right]{
+            return false
+        }
+        //移动左右指针
+        left++
+        right--
+    }
+    return true
+}
+```
 
 javaScript：