From c46acca1e4ceaddbda6b4975afe0c4d061ab11fc Mon Sep 17 00:00:00 2001
From: ArthurP <gpan20020205@gmail.com>
Date: Sun, 14 Nov 2021 10:05:29 +0000
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---
 ...除字符串中的所有相邻重复项.md | 51 +++++++++++++++++++
 1 file changed, 51 insertions(+)

diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md
index f70f39f3..ef93ced8 100644
--- a/problems/1047.删除字符串中的所有相邻重复项.md
+++ b/problems/1047.删除字符串中的所有相邻重复项.md
@@ -269,6 +269,57 @@ var removeDuplicates = function(s) {
 };
 ```
 
+C:
+方法一：使用栈
+```c
+char * removeDuplicates(char * s){
+    //求出字符串长度
+    int strLength = strlen(s);
+    //开辟栈空间。栈空间长度应为字符串长度+1（为了存放字符串结束标志'\0'）
+    char* stack = (char*)malloc(sizeof(char) * strLength + 1);
+    int stackTop = 0;
+
+    int index = 0;
+    //遍历整个字符串
+    while(index < strLength) {
+        //取出当前index对应字母，之后index+1
+        char letter = s[index++];
+        //若栈中有元素，且栈顶字母等于当前字母（两字母相邻）。将栈顶元素弹出
+        if(stackTop > 0 && letter == stack[stackTop - 1])
+            stackTop--;
+        //否则将字母入栈
+        else
+            stack[stackTop++] = letter;
+    }
+    //存放字符串结束标志'\0'
+    stack[stackTop] = '\0';
+    //返回栈本身作为字符串
+    return stack;
+}
+```
+方法二：双指针法
+```c
+char * removeDuplicates(char * s){
+    //创建快慢指针
+    int fast = 0;
+    int slow = 0;
+    //求出字符串长度
+    int strLength = strlen(s);
+    //遍历字符串
+    while(fast < strLength) {
+        //将当前slow指向字符改为fast指向字符。fast指针+1
+        char letter = s[slow] = s[fast++];
+        //若慢指针大于0，且慢指针指向元素等于字符串中前一位元素，删除慢指针指向当前元素
+        if(slow > 0 && letter == s[slow - 1])
+            slow--;
+        else
+            slow++;
+    }
+    //在字符串结束加入字符串结束标志'\0'
+    s[slow] = 0;
+    return s;
+}
+```
 
 
 -----------------------