LeetCode-42 Trapping Rain Water

时间：2015-02-17 07:01:18 收藏：0 阅读：231

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.

技术分享

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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Array Stack Two Pointers

注意：题目给出的图片具有很大迷惑性。需要注意的是，即使是局部的最高点，在全局看来，也有可能储水。如一个case{6,4,2,0,3,2,0,3,1,4,5}.

解题思路：

1. 先找到全局的最高点；

2. 分别从左右两边像最高点逼近。

3. 由左边向最高点逼近时，设左边当前的最大值下标为currentMax,当前遍历到i，全局最高值下边为maxIndex;

　如果A[currentMax] > A[i]; sum += (A[currentMax] - A[i]);

　否则，currentMax = i，此条带无法储水；

4. 由右边逼近最高点类似3.

5. 算法时间复杂度为O(n)

代码如下：

public int trap(int[] A) {
        int maxIndex = 0;
        for(int i=1; i<A.length; i++) {
            if(A[i] > A[maxIndex])
                maxIndex = i;
        }
        
        int sum = 0;
        //from left to top
        int currentMax = 0;
        for(int i=1; i<maxIndex; i++) {
            if(A[i] < A[currentMax])
                sum += (A[currentMax] - A[i]);
            else 
                currentMax = i;
        }
                //from right to top
        currentMax = A.length-1;
        for(int i=A.length-2; i>maxIndex; i--) {
            if(A[i] < A[currentMax])
                sum += (A[currentMax] - A[i]);
            else
                currentMax = i;
        }
        
        return sum;
    }

原文：http://www.cnblogs.com/linxiong/p/4294777.html