[leetcode]42. Trapping Rain Water雨水积水问题

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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

题意：

给定一个地形，计算能存多少雨水。

思路：

扫数组，找到最高柱子并以此为中心，将数组分为两半。 

对左半部分而言，可以想象右边一定有堵墙，只需要每次更新leftMost就能决定water大小

对右半部分而言，同理。

代码：

 1 class Solution {
 2     public int trap(int[] height) {
 3         //1.find the peak_ index
 4         int peak_index = 0;
 5         for(int i = 0; i< height.length; i++){
 6             if(height[i] > height[peak_index]){
 7                 peak_index=i;
 8             }
 9         }
10         //2.left --> right, find the leftMost, check current height < leftMost ? water , update leftMost 
11         int leftMost = 0;
12         int water = 0;
13         for(int i = 0; i < peak_index; i++){
14             if(height[i]>leftMost){
15                 leftMost = height[i];
16             }else{
17                 water = water +  leftMost -  height[i];
18             }  
19         }
20         //3.right --> left
21         int rightMost = 0;
22         for(int i = height.length-1; i>peak_index; i--){
23             if(height[i]>rightMost){
24                 rightMost = height[i];
25             }else{
26                 water = water +  rightMost -  height[i];
27             }  
28         }
29         return water;     
30     }
31 }