LeetCode-42 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!

 

Hide Tags

 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{5,2,1,2,1,5}.

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[] height) {
        if(height.length == 0) return 0;
        int maxIndex = 0;
        for(int i=0; i<height.length; i++) maxIndex = (height[maxIndex] > height[i] ? maxIndex : i);
        
        int sum = 0;
        //from left to right
        int currMax = 0;
        for(int i=0; i<maxIndex; i++) {
            if(height[i] < height[currMax]) sum += (height[currMax] - height[i]);
            else currMax = i;
        }
        //from right to left 
        currMax = height.length-1;
        for(int i=height.length-1; i>maxIndex; i--) {
            if(height[i] < height[currMax]) sum += (height[currMax] - height[i]);
            else currMax = i;
        }
        return sum;
    }