[Leetcode] Trapping Rain Water (Java)

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example, 
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

给定数组A[]，数组中的每个数代表墙的高度，A[]就代表一组墙的形状，向这组墙中蓄水，问这组墙能存放多少水。

思路：

1）找到最高点，然后分成左、右两部分；

2）在右半边中，从左向右找第一个最高点（最高点可能有好几个），计算这之中存水，然后继续2）直到最右边

3）在左半边中，从右向左找第一个最高点（最高点可能有好几个），计算这之中存水，然后继续3）直到最左边

public class TrappingRainWater {public int trap(int[] A) {int res = 0;int max = 0;int index = 0;for(int i=0;i<A.length;i++){if(A[i]>max){max=A[i];index = i;}}int pre = index+1;while(pre<A.length){int rMax = A[pre];int rIndex = pre; for(int i=pre;i<A.length;i++){if(A[i]>rMax){rMax=A[i];rIndex=i;}}for(int i=pre;i<rIndex;i++){res+=rMax-A[i];}pre = rIndex+1;}int post = index-1;while(post>0){int lMax = A[post];int lIndex = post; for(int i=post;i>=0;i--){if(A[i]>lMax){lMax=A[i];lIndex=i;}}for(int i=post;i>lIndex;i--){res+=lMax-A[i];}post = lIndex -1;}return res;}public static void main(String[] args) {int[] A = {0,1,0,2,1,0,1,3,2,1,2,1};System.out.println(new TrappingRainWater().trap(A));}}