今日學(xué)習(xí)
不同路徑
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https://programmercarl.com/0062.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84.html
視頻講解：https://www.bilibili.com/video/BV1ve4y1x7Eu
第一想法
動(dòng)態(tài)規(guī)劃的思想能岩，先定義第一行和第一列，然后每個(gè)dp[i][j]都根據(jù)dp[i-1][j]+dp[i][j-1]來(lái)計(jì)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. 不同路徑 II
視頻講解：https://www.bilibili.com/video/BV1Ld4y1k7c6
第一想法
主要是遇到障礙物昔头，后續(xù)的賦值就直接為0伊群，記住這個(gè)定義二維數(shù)組的寫(xiě)法 const dp = Array(m).fill().map(item => Array(n).fill(0));
先要獲取這個(gè)二維數(shù)組的長(zhǎng)寬考杉，初始化的時(shí)候遇到障礙物就停掉可以用break也可以將obstacleGrid[i][0] === 0寫(xiě)在for循環(huán)條件判斷中
/**
* @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]
};
