【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!

思路：

用两个数组，leftMax[i]表示左边到i-1为之最高的高度，rightMax同上。

某位置能存储的水量等于min{leftMax[i],rightMax[i]}-height[i]

算法：

public int trap(int[] height) {  
    int water = 0;  
    int left_max[] = new int[height.length];  
    int right_max[] = new int[height.length];  
    for (int j = 1; j < height.length - 1; j++) {  
        left_max[j] = Math.max(left_max[j - 1], height[j - 1]);  
    }  
    for (int j = height.length - 2; j >=0; j--) {  
        right_max[j] = Math.max(right_max[j + 1], height[j + 1]);  
    }  
    for (int i = 1; i < height.length - 1; i++) {  
        int tmp = Math.min(left_max[i], right_max[i]) - height[i];  
        if (tmp > 0)  
            water += tmp;  
    }  
    return water;  
}  

【Leetcode】Trapping Rain Water

原文：http://blog.csdn.net/yeqiuzs/article/details/51510886