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Candy

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?

思路:分糖果问题,用贪心算法。我们用num[i]表示第i个孩子的糖果数目,如果i+1孩子比i孩子的ratings大,则i+1的孩子糖果数目为i孩子糖果数目+1.如果i孩子比i+1孩子ratings大,怎么办,等从左往右开始遍历结束,我们从右往左开始,把之前忽略的情况给补上,就可以了。时间复杂度为O(n),空间复杂度为O(n)。

class Solution {public:    int candy(vector<int> &ratings) {        int n=ratings.size();        if(n<=0)            return 0;        vector<int> num(n,1);        for(int i=1;i<n;i++)        {            if(ratings[i-1]<ratings[i])            {                num[i]=num[i-1]+1;            }        }        for(int i=n-2;i>=0;i--)        {            if(ratings[i]>ratings[i+1] && num[i]<=num[i+1])            {                num[i]=num[i+1]+1;            }        }        int sum=0;        for(int i=0;i<n;i++)        {            sum+=num[i];        }        return sum;    }};