【leetcode】Trapping Rain Water

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!

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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：

1. 首先找出最大的bar，索引记录在totalMaxIndex这个变量里面；
2. 从左往右，依次处理totalMaxIndex左边的bar；设置一个curMax记录遍历到当前为止最大的bar，那么对于A[i]，有两种可能：一是A[i]<curMax，那么这时候就可以储水(curMax - A[i])；而是A[i] > curMax，这时更新curMax = A[i]；
3. 从右往左，依次处理totalMaxIndex右边的bar，方法同2.

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：

public class Solution {
    public int trap(int[] A) {
        if(A == null || A.length == 0)
            return 0;
        
        int curMax = 0;
        int totalMax = A[0];
        int totalMaxIndex = 0;
        int answer = 0;
        
        for(int i = 1;i < A.length;i++){
            if(A[i] > totalMax){
                totalMax = A[i];
                totalMaxIndex = i;
            }
        }
        
        for(int i = 0;i < totalMaxIndex;i++){
            if(A[i] < curMax)
                answer += curMax - A[i];
            else
                curMax = A[i];
        }
        
        curMax = 0;
        for(int i = A.length-1;i > totalMaxIndex;i--){
            if(A[i] < curMax)
                answer += curMax - A[i];
            else {
                curMax = A[i];
            }
        }
        
        return answer;
    }
}