今日学习

不同路径

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https://programmercarl.com/0062.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84.html

视频讲解：https://www.bilibili.com/video/BV1ve4y1x7Eu

第一想法

动态规划的思想，先定义第一行和第一列，然后每个dp[i][j]都根据dp[i-1][j]+dp[i][j-1]来计算出

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function (m, n) {
    let dp = new Array(m).fill().map(item => Array(n).fill(0))
    for (let i = 0; i < m; i++) {
        dp[i][0] = 1
    }
    for (let j = 0; j < n; j++) {
        dp[0][j] = 1
    }
    for (let i = 1; i < m; i++) {
        for (let j = 1; j < n; j++) {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
        }
    }
    return dp[m - 1][n - 1]
};

63. 不同路径 II

https://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.htmlhttps://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.html

视频讲解：https://www.bilibili.com/video/BV1Ld4y1k7c6

第一想法

主要是遇到障碍物，后续的赋值就直接为0，记住这个定义二维数组的写法 const dp = Array(m).fill().map(item => Array(n).fill(0));
先要获取这个二维数组的长宽，初始化的时候遇到障碍物就停掉可以用break也可以将obstacleGrid[i][0] === 0写在for循环条件判断中

/**
 * @param {number[][]} obstacleGrid
 * @return {number}
 */
var uniquePathsWithObstacles = function (obstacleGrid) {
    const m = obstacleGrid.length
    const n = obstacleGrid[0].length

    const dp = Array(m).fill().map(item => Array(n).fill(0));
    for (let i = 0; i < m && obstacleGrid[i][0] === 0; i++) {
        dp[i][0] = 1
    }
    for (let j = 0; j < n && obstacleGrid[0][j] === 0; j++) {
        dp[0][j] = 1
    }
    for (let i = 1; i < m; i++) {
        for (let j = 1; j < n; j++) {
            if (obstacleGrid[i][j]) {
                dp[i][j] = 0
            } else {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
            }
        }
    }
    return dp[m - 1][n - 1]
};