42. 接雨水 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 can trap after raining.

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) 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.

 

方法一：

暴力法，依次计算并分别查找左边最高和右边最高

 

方法二：

先计算出左边最高和右边最高，分别放入数组，再依次计算

 

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

 

参考链接：

https://leetcode.com/problems/trapping-rain-water/

https://leetcode-cn.com/problems/trapping-rain-water/