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!

 

标签 Array Stack Two Pointers

类似的题目 (M) Container With Most Water (M) Product of Array Except Self (H) Trapping Rain Water II

 

 

 

思路：两头指针，一个柱体一个柱体遍历，有可以积水的直接加。对于[left,right]，height[left]表示的是“碗”左边高度，height[right]表示的是“碗”右边高度，level表示的是当前“碗”的最高盛水高度，lower=min(height[left]，height[right])。

 

代码：

public class Solution {
    public int trap(int[] height) {
        int water = 0, lower = 0, level = 0, left = 0, right = height.length - 1;
        while(left < right) {
            lower = height[height[left] <= height[right] ? left++ : right--];
            level = Math.max(level, lower);
            water += level - lower;
        }
        return water;
    }
}