[leedcode 135] Candy

標籤：

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) {        /*其實只要左右不比他大，就可以只分1個。如果左右比他大，左右就多分一個。這樣的結果就是最省的。                其實題目的描述本身就是一個很好的方案：初始化將每個人的糖果數都初始化為1。每次遍曆只考慮左右一邊，        即從左向右遍曆，如果i>i-1 則  candy[i]=candy[i-1]+1；再從右向左遍曆一次，如果i>i+1 並且 candy[i]<=candy[i+1](注意！！)         則 cand[i]=candy[i+1]+1; 這樣的兩次遍曆，左右兩邊都顧及到了。最後將candy[i]相加求和就好了。        每一邊只考慮遞增序列！！！*/        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;            }}

 

[leedcode 135] Candy