@@ -238,6 +238,47 @@ var candy = function(ratings) {
238238};
239239```
240240
241+ ### C
242+ ``` c
243+ #define max (a, b ) (((a) > (b)) ? (a) : (b))
244+ 245+ int *initCandyArr (int size) {
246+ int * candyArr = (int* )malloc(sizeof(int) * size);
247+ 248+ int i;
249+ for(i = 0; i < size; ++i)
250+ candyArr[i] = 1;
251+ 252+ return candyArr;
253+ }
254+ 255+ int candy(int* ratings, int ratingsSize){
256+ // 初始化数组,每个小孩开始至少有一颗糖
257+ int * candyArr = initCandyArr(ratingsSize);
258+ 259+ int i;
260+ // 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)
261+ for(i = 1; i < ratingsSize; ++i) {
262+ if(ratings[i] > ratings[i - 1])
263+ candyArr[i] = candyArr[i - 1] + 1;
264+ }
265+ 266+ // 再判断左边评分是否比右边高。
267+ // 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)
268+ // 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]
269+ for(i = ratingsSize - 2; i >= 0; --i) {
270+ if(ratings[i] > ratings[i + 1])
271+ candyArr[i] = max(candyArr[i], candyArr[i + 1] + 1);
272+ }
273+ 274+ // 求出糖果之和
275+ int result = 0;
276+ for(i = 0; i < ratingsSize; ++i) {
277+ result += candyArr[i];
278+ }
279+ return result;
280+ }
281+ ```
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243284<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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