第十一周 leetcode 135. Candy（Hard）

题目描述：

There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
  Each child must have at least one candy.
  Children with a higher rating get more candies than their neighbors.
  What is the minimum candies you must give?

解题思路：
这题我们用贪心算法，求出局部最优解，最后可以得到整体最优解。
1.与前面的邻居比较，前向遍历权重数组ratings，如果ratings[i]>ratings[i-1]，则cost[i]=cost[i-1]+1；
2.与后面的邻居比较，后向遍历权重数组ratings，如果ratings[i]>ratings[i+1]且 cost[i]小于cost[i+1]+1，则更新cost，cost[i]=cost[i+1]+1；
3、对cost求和即为最少需要的糖果。

代码：

class Solution {public:    int candy(vector<int>& ratings) {        int n=ratings.size();        int cost[n];        for (int i = 0; i < n; i++) cost[i] = 1;        // 向后遍历        for(int i = 1;i < n; i++) {            if(ratings[i] > ratings[i-1])                cost[i] = cost[i-1] + 1;        }        // 向前遍历        int res = cost[n-1];        for(int i = n-2; i >= 0; i--) {            if(ratings[i] > ratings[i+1] && cost[i] < cost[i+1]+1)                cost[i] = cost[i+1] + 1;            res += cost[i];        }        return res;    }};

代码运行结果：