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原题链接在这里：https://leetcode.com/problems/candy/
与Trapping Rain Water相似。每个孩子能拿到多少糖取决于左右两边，和能储存多少水取决于Math.min(左侧挡板最大值, 右侧挡板最大值)一个道理。
所以先用leftNum保存根据左侧邻居，该孩子能拿到多少糖，若是rating比左侧高，就要比左侧多拿一个，否则只需要拿一个。右侧同理。
最后在走一遍，取左右侧中较大值加入res中。
Time O(n), Space O(n).
AC Java:
1 public class Solution { 2 public int candy(int[] ratings) { 3 if(ratings == null || ratings.length == 0){ 4 return 0; 5 } 6 7 //get how many candies based on left neighbour 8 int [] leftNum = new int[ratings.length]; 9 leftNum[0] = 1; 10 for(int i = 1; i<ratings.length; i++){ 11 if(ratings[i] > ratings[i-1]){ 12 leftNum[i] = leftNum[i-1]+1; 13 }else{ 14 leftNum[i] = 1; 15 } 16 } 17 18 //get how many candies base on right neighbour 19 int [] rightNum = new int[ratings.length]; 20 rightNum[ratings.length-1] = 1; 21 for(int i = ratings.length-2; i>=0; i--){ 22 if(ratings[i] > ratings[i+1]){ 23 rightNum[i] = rightNum[i+1]+1; 24 }else{ 25 rightNum[i] = 1; 26 } 27 } 28 29 //total number of candies 30 int res = 0; 31 for(int i = 0; i<ratings.length; i++){ 32 res += Math.max(leftNum[i], rightNum[i]); 33 } 34 return res; 35 } 36 }
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/4938112.html
