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;
}