【leetcode】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?

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题解：

1. 第一个孩子给一颗糖，然后从左到右遍历rate数组，如果孩子i+1的rate比孩子i的rate高，那么孩子i+1得到的糖果数目比孩子i得到的糖果数目多1；如果孩子i+1的rate比孩子i的rate低，那么就给孩子i+1一颗糖；
2. 从右往左遍历rate数组，如果当前遍历的孩子i的rate比他右边的孩子的rate高，那么他得到的糖果就比他右边的孩子得到的糖果多1.
3. 累加rate中所有的值，得到总的最少糖果数目。

代码如下：

public class Solution {
    public int candy(int[] ratings) {
        if(ratings.length == 0)
            return 0;
        
        int[] count = new int[ratings.length];
        Arrays.fill(count, 1);
        
        for(int i = 1;i <= ratings.length-1;i++){
            if(ratings[i]> ratings[i-1] )
                count[i] = count[i-1] + 1; 
        }
        
        int sum = 0;
        for(int i = ratings.length-1;i >= 1;i--){
            sum += count[i];
            if(ratings[i-1] > ratings[i] && count[i-1] <= count[i])
                count[i-1] = count[i]+ 1; 
        }
        
        return count[0] + sum;
    }
}

在17行的循环，需要判断i-1个孩子当前获得的糖果数目是否真的比第i个孩子的少，如果真的少，才需要+1.例如：

ratings = {4,2,3,4,1}的时候，第一遍遍历得到的count数组是{1,1,2,3,1}，此时从后往前遍历的时候ratings[3] > ratings[4]，但是count[3]已经大于count[4]了，所以不需要更新count[3] = count[4]+1。