From 84f41f6685e744b138890cf7143ed3069a02a78e Mon Sep 17 00:00:00 2001
From: Guanzhong Pan <gpan20020205@gmail.com>
Date: Tue, 19 Apr 2022 17:00:17 +0100
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200135.=E5=88=86=E5=8F=91?=
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---
 problems/0135.分发糖果.md | 41 +++++++++++++++++++++++++++++++++++
 1 file changed, 41 insertions(+)

diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md
index ccdabc16..27071aee 100644
--- a/problems/0135.分发糖果.md
+++ b/problems/0135.分发糖果.md
@@ -238,6 +238,47 @@ var candy = function(ratings) {
 };
 ```
 
+### C
+```c
+#define max(a, b) (((a) > (b)) ? (a) : (b))
+
+int *initCandyArr(int size) {
+    int *candyArr = (int*)malloc(sizeof(int) * size);
+
+    int i;
+    for(i = 0; i < size; ++i)
+        candyArr[i] = 1;
+
+    return candyArr;
+}
+
+int candy(int* ratings, int ratingsSize){
+    // 初始化数组，每个小孩开始至少有一颗糖
+    int *candyArr = initCandyArr(ratingsSize);
+
+    int i;
+    // 先判断右边是否比左边评分高。若是，右边孩子的糖果为左边孩子+1（candyArr[i] = candyArr[i - 1] + 1)
+    for(i = 1; i < ratingsSize; ++i) {
+        if(ratings[i] > ratings[i - 1])
+            candyArr[i] = candyArr[i - 1] + 1;
+    }
+
+    // 再判断左边评分是否比右边高。
+    // 若是，左边孩子糖果为右边孩子糖果+1/自己所持糖果最大值。（若糖果已经比右孩子+1多，则不需要更多糖果）
+    // 举例：ratings为[1, 2, 3, 1]。此时评分为3的孩子在判断右边比左边大后为3，虽然它比最末尾的1(ratings[3])大，但是candyArr[3]为1。所以不必更新candyArr[2]
+    for(i = ratingsSize - 2; i >= 0; --i) {
+        if(ratings[i] > ratings[i + 1])
+            candyArr[i] = max(candyArr[i], candyArr[i + 1] + 1);
+    }
+
+    // 求出糖果之和
+    int result = 0;
+    for(i = 0; i < ratingsSize; ++i) {
+        result += candyArr[i];
+    }
+    return result;
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>