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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!
题解：在网上研究了半天，才找到时间O(n)和空间O(1)的算法：
具体算法过程如下图所示：
基本的原理就是当从左往右遍历的时候，最高的bar保证了能够在右端将水拦住，那么就只要计算对于A[i]，在左端能够拦截住多高的水就可以了，即变量curMax。
代码如下：
1 public class Solution { 2 public int trap(int[] A) { 3 if(A == null || A.length == 0) 4 return 0; 5 6 int curMax = 0; 7 int totalMax = A[0]; 8 int totalMaxIndex = 0; 9 int answer = 0; 10 11 for(int i = 1;i < A.length;i++){ 12 if(A[i] > totalMax){ 13 totalMax = A[i]; 14 totalMaxIndex = i; 15 } 16 } 17 18 for(int i = 0;i < totalMaxIndex;i++){ 19 if(A[i] < curMax) 20 answer += curMax - A[i]; 21 else 22 curMax = A[i]; 23 } 24 25 curMax = 0; 26 for(int i = A.length-1;i > totalMaxIndex;i--){ 27 if(A[i] < curMax) 28 answer += curMax - A[i]; 29 else { 30 curMax = A[i]; 31 } 32 } 33 34 return answer; 35 } 36 }
【leetcode】Trapping Rain Water,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/sunshineatnoon/p/3853029.html
