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.

package com.bupt.acm.leetcode;

import java.util.Arrays;

public class Candy {

    /**
     * 本算用于求解当等级比其高时值加倍
     * @param ratings
     * @return
     */
    public int sum=0;
    public int candy(int[] ratings){
        
        if(ratings==null||ratings.length==0){
            return 0;
        }
        int min=ratings[0];
        int length=ratings.length;
        if(ratings.length==1){
            return 1;
        }
        
        int[] score=new int[length];
        for(int i=0;i<length;i++)
            score[i]=0;
        for(int i=0;i<length;i++){
            if(min>ratings[i]){
                min=ratings[i];
            }
        }
        for(int i=0;i<length;i++){
            if(ratings[i]==min){
                score[i]=1;
            }else{
                score[i]=computeCandy(ratings,i-1,i,i+1,score);
            }
        }
        System.out.println("result");
        for(int i=0;i<length;i++){
            sum+=score[i];
        }
        return sum;
    }
    private int computeCandy(int[] ratings,int beg,int now,int end,int[] score) {
        // TODO Auto-generated method stub
        if(score[now]!=0)
            return score[now];
        if(now==ratings.length-1){//now:表示最后一个元素
            if(ratings[beg]>=ratings[now]){
                return 1;
            }else{
                return computeCandy(ratings, beg-1, beg, now,score)+1;
            }
        }else if(now==0){//表示第一个元素
            if(ratings[now]<=ratings[end])
                return 1;
            else{
                return computeCandy(ratings, now, end, end+1,score)+1;
            }
        }else{//表示其他元素
            int a=ratings[beg],b=ratings[now],c=ratings[end];
            if(a==b&&b==c){//a=b=c
                return 1;
            }else if(b>a&&b>c){//a<b>c
                return Math.max(computeCandy(ratings, beg-1, beg, now,score)+1, computeCandy(ratings, now, end, end+1,score)+1);
            }else if(b>a&&b<=c){// a<b<=c
                return computeCandy(ratings, beg-1, beg, now,score)+1;
            }else if(b<=a&&b>c){//a<b<c
                return computeCandy(ratings, now, end, end+1,score)+1;
            }else{//b<=a&&b<=c
                return 1;
            }
        }
    }

    public static void main(String[] args) {
        int[] ratings={1,4,5,2,7,2,5};
        Candy candy=new Candy();
        System.out.println(candy.candy(ratings));
    }
}
//方法二算法：

 上述算法最后一个样例超时，

新算法：

初始化所有小孩糖数目为1，从前往后扫描，如果第i个小孩等级比第i-1个高，那么i的糖数目等于i-1的糖数目+1；从后往前扫描，如果第i个的小孩的等级比i+1个小孩高,但是糖的数目却小或者相等，那么i的糖数目等于i+1的糖数目+1

    public int candy2(int[] ratings){
        if(ratings==null||ratings.length==0)
            return 0;
        if(ratings.length==1)
            return 1;
        int sum=0;
        int length=ratings.length;
        int[] scores=new int[length];
        for(int i=0;i<length;i++){
            scores[i]=1;
        }
        for(int i=1;i<length;i++){
            if(ratings[i]>ratings[i-1]){
                scores[i]=scores[i-1]+1;
            }
        }
        for(int i=length-1;i>0;i--){
            if(ratings[i-1]>ratings[i]){
                if(scores[i-1]<=scores[i]){
                    scores[i-1]=scores[i]+1;
                }
            }
        }
        for(int i=0;i<length;i++){
            sum+=scores[i];
        }
        return sum;
    }