【LeetCode】 42. Trapping Rain Water 接雨水（Hard）（JAVA）

【LeetCode】 42. Trapping Rain Water 接雨水（Hard）（JAVA）

题目地址: https://leetcode.com/problems/trapping-rain-water/

题目描述：

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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

题目大意

给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。

解题方法

1、计算出从左往右的最大值
2、计算出从右往左的最大值
3、判断上面两个最大值是否比当前值大

class Solution {
    public int trap(int[] height) {
        if (height.length <= 2) return 0;
        int[] lefts = new int[height.length];
        int[] rights = new int[height.length];
        for (int i = 0; i < height.length; i++) {
            if (i == 0) {
                lefts[i] = height[i];
                rights[height.length - 1 - i] = height[height.length - 1 - i];
            } else {
                lefts[i] = Math.max(height[i], lefts[i - 1]);
                rights[height.length - 1 - i] = Math.max(height[height.length - 1 - i], rights[height.length - i]);
            }
        }
        int res = 0;
        for (int i = 0; i < height.length; i++) {
            if (height[i] >= lefts[i] || height[i] >= rights[i]) continue;
            res += Math.min(lefts[i], rights[i]) - height[i];
        }
        return res;
    }
}

执行用时 : 2 ms, 在所有 Java 提交中击败了 47.67% 的用户
内存消耗 : 38.6 MB, 在所有 Java 提交中击败了 9.45% 的用户