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
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]
};
