LeetCode每日一练【11】

LeetCode每日一练 -- Container With Most Water

Brute Force

/*
 * @Author: fox
 * @Date: 2022-04-27 09:07:49
 * @LastEditors: fox
 * @LastEditTime: 2022-04-27 11:02:06
 * @Description: https://leetcode.com/problems/container-with-most-water/
 */

/**
 * @description: Brute Force
 * 虽然成功了，但是...Time Limit Exceeded！在面对大量数据的时候，这个方法显然是不合适的，效率太低
 * @param {number[]} height 
 * @return {number}
 */
const maxArea = (height) => {
    // 1. 如果height数组的长度为0或1的时候，直接返回0
    if (!height || !height.length || height.length === 1) return 0;
    
    let area = 0; // 最大区域

    // 2. 双重循环遍历，计算所有区域面积，然后取最大值
    for (let x = 0; x < height.length; x++) {
        for (let y = x + 1; y < height.length; y++) {
            const small = height[x] < height[y] ? height[x] : height[y];
            const tempArea = small * (y - x);
            area = tempArea > area ? tempArea : area;
        };
    };
    return area
};

let height;

height = [1, 8, 6, 2, 5, 4, 8, 3, 7];
console.log(maxArea(height)); // 49

height = [1, 1];
console.log(maxArea(height)); // 1

Two Pointer Approach

/*
 * @Author: fox
 * @Date: 2022-04-27 09:07:49
 * @LastEditors: fox
 * @LastEditTime: 2022-04-27 16:06:55
 * @Description: https://leetcode.com/problems/container-with-most-water/
 */

/**
 * @description Two Pointer Approach
 * @param {number[]} height
 * @return {number}
 */
const maxArea = (heights) => {
    let [l, r] = [0, heights.length - 1]
    let max = 0;
    
    while (l < r) {
        max = Math.max(max, (r - l) * Math.min(heights[l], heights[r]));
          
        // 初始状态下，x轴的宽度我们已经获取了最大值，并且担任了最大区域，这个时候如果不移动较短的一条线，区域并不会再增加，并且
        // 还会减少，这是因为height取决于最短的一条线。所以应该移动它，这样矩形区域在x轴的损失的面积关联可能会被高度弥补
        if (heights[l] < heights[r]) {
            l++;
        } else {
            r++;
        }
    }
    
  return max;
};

let height;

height = [1, 8, 6, 2, 5, 4, 8, 3, 7];
console.log(maxArea(height)); // 49

height = [1, 1];
console.log(maxArea(height)); // 1