[leedcode 135] 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?

public class Solution {
    public int candy(int[] ratings) {
        /*其实只要左右不比他大，就可以只分1个。如果左右比他大，左右就多分一个。这样的结果就是最省的。

        其实题目的描述本身就是一个很好的方案：初始化将每个人的糖果数都初始化为1。每次遍历只考虑左右一边，
        即从左向右遍历，如果i>i-1 则  candy[i]=candy[i-1]+1；再从右向左遍历一次，如果i>i+1 并且 candy[i]<=candy[i+1](注意！！)
        则 cand[i]=candy[i+1]+1; 这样的两次遍历，左右两边都顾及到了。最后将candy[i]相加求和就好了。
        每一边只考虑递增序列！！！*/
        if(ratings==null||ratings.length<1) return 0;
        int len=ratings.length;
        int candy[]=new int[len];
        int res=0;
        for(int i=0;i<len;i++){
            candy[i]=1;
        }
        for(int i=1;i<len;i++){
            if(ratings[i-1]<ratings[i]){
                candy[i]=candy[i-1]+1;
            }

        }
        for(int i=len-2;i>=0;i--){
            if(ratings[i+1]<ratings[i]&&candy[i]<=candy[i+1]){////
                candy[i]=candy[i+1]+1;
            }
        }
        for(int i=0;i<len;i++){
            res+=candy[i];
        }
        return res;

    }
}