LeetCode 42. Trapping Rain Water

最新推荐文章于 2023-08-07 09:17:41 发布

原创最新推荐文章于 2023-08-07 09:17:41 发布 · 813 阅读

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两个指针同时被 3 个专栏收录

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容器

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两个最值

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本文针对LeetCode上的雨水收集问题提供了一种高效的解决方案。通过使用双指针法，该算法能够计算出给定高度数组中能收集到的雨水总量。文章详细介绍了如何利用左右两侧的最大值来确定每个位置上能收集的水量。

原题网址：https://leetcode.com/problems/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!

方法：设置两个夹逼指针，两个最值指针。

public class Solution {
    public int trap(int[] height) {
        if (height == null || height.length <= 2) return 0;
        int sum = 0;
        int i = 0, j = height.length-1;
        int left = height[0], right = height[j];
        while (i < j) {
            if (left <= right) {
                i ++;
                left = Math.max(left, height[i]);
                sum += left-height[i];
            } else {
                j --;
                right = Math.max(right, height[j]);
                sum += right-height[j];
            }
        }
        return sum;
    }
}