62. Unique Paths

https://leetcode.com/problems/unique-paths/

注意本題只能向右 / 向上走。

DP 問題，經(jīng)典又熟悉。

暴力遞歸->傻緩存->動(dòng)態(tài)規(guī)劃。

class Solution {int M, N;public int uniquePaths(int m, int n) {M = m;N = n;// Approach 1: Recursive, Brute Force // return process1(0, 0);// Approach 2: Recursion with Memoization // int[][] dp = new int[M][N]; // dp[M - 1][N - 1] = 1; // return process2(0, 0, dp);// Approach 3: Dynamic Programmingint[][] dp = new int[M][N];for (int i = M - 1; i >= 0; i--) {dp[i][N - 1] = 1;}for (int j = N - 1; j >= 0; j--) {dp[M - 1][j] = 1;}for (int i = M - 2; i >= 0; i--) {for (int j = N - 2; j >= 0; j--) {dp[i][j] = dp[i + 1][j] + dp[i][j + 1];}}return dp[0][0];}// public int process2(int i, int j, int[][] dp) { // 從i,j到M,N共有多少種路徑 // if (i == M || j == N) return 0; // if (dp[i][j] > 0) return dp[i][j]; // dp[i][j] = process2(i + 1, j, dp) + process2(i, j + 1, dp); // return dp[i][j]; // }// public int process1(int i, int j) { // if (i == M || j == N) return 0; // if (i == M - 1 && j == N - 1) return 1; // return process1(i + 1, j) + process1(i, j + 1); // } }

63. Unique Paths II

https://leetcode.com/problems/unique-paths-ii/

DP 問題，同上題，注意本題只能向右 / 向上走，只不過多了一些障礙物。
暴力遞歸->傻緩存->動(dòng)態(tài)規(guī)劃。

class Solution {int M, N;public int uniquePathsWithObstacles(int[][] grid) {M = grid.length;N = grid[0].length;// Approach 1: Recursive, Brute Force // return process1(0, 0, grid);// Approach 2: Recursion with Memoization// int[][] dp = new int[M][N];// for (int i = 0; i < M; i++) {// Arrays.fill(dp[i], -1);// }// dp[M - 1][N - 1] = 1 - grid[M - 1][N - 1];// for (int i = M - 2; i >= 0; i--) {// if (grid[i][N - 1] == 1 || dp[i + 1][N - 1] == 0) dp[i][N - 1] = 0;// else dp[i][N - 1] = 1;// }// for (int j = N - 2; j >= 0; j--) {// if (grid[M - 1][j] == 1 || dp[M - 1][j + 1] == 0) dp[M - 1][j] = 0;// else dp[M - 1][j] = 1;// }// return process2(0, 0, grid, dp);// Approach 3: Dynamic Programmingint[][] dp = new int[M][N];dp[M - 1][N - 1] = 1 - grid[M - 1][N - 1];for (int i = M - 2; i >= 0; i--) {if (grid[i][N - 1] == 1 || dp[i + 1][N - 1] == 0) dp[i][N - 1] = 0;else dp[i][N - 1] = 1;}for (int j = N - 2; j >= 0; j--) {if (grid[M - 1][j] == 1 || dp[M - 1][j + 1] == 0) dp[M - 1][j] = 0;else dp[M - 1][j] = 1;}for (int i = M - 2; i >= 0; i--) {for (int j = N - 2; j >= 0; j--) {if (grid[i][j] == 1) dp[i][j] = 0;else dp[i][j] = dp[i + 1][j] + dp[i][j + 1];}}return dp[0][0];}// public int process2(int i, int j, int[][] grid, int[][] dp) { // 從i,j到M,N共有多少種路徑 // if (i == M || j == N || grid[i][j] == 1) return 0; // if (dp[i][j] >= 0) return dp[i][j]; // dp[i][j] = process2(i + 1, j, grid, dp) + process2(i, j + 1, grid, dp); // return dp[i][j]; // }// public int process1(int i, int j, int[][] grid) { // if (i == M || j == N || grid[i][j] == 1) return 0; // if (i == M - 1 && j == N - 1) return 1; // return process1(i + 1, j, grid) + process1(i, j + 1, grid); // } }

980. Unique Paths III

https://leetcode.com/problems/unique-paths-iii/

本題是有史以來(lái)做過的最簡(jiǎn)單的 hard 了，就是一個(gè)沒有任何優(yōu)化技巧的 backtracing，30分鐘搞定。

思路如下：因?yàn)?exactly once 語(yǔ)義，所以不能走重復(fù)路線，故維護(hù)一個(gè) visited 數(shù)組。另外維護(hù)一個(gè) remain 代表剩余步數(shù)，當(dāng)經(jīng)過終點(diǎn)時(shí)，看剩余步數(shù)是否恰好為0，或者當(dāng)剩余步數(shù)為 0 時(shí)，看是否恰好經(jīng)過終點(diǎn)。否則沒有必要繼續(xù)走下去。

class Solution {int M, N;int startX, startY;int endX, endY;public int uniquePathsIII(int[][] grid) {M = grid.length;N = grid[0].length;int remain = M * N;for (int i = 0; i < M; i++) {for (int j = 0; j < N; j++) {if (grid[i][j] == -1) remain--;if (grid[i][j] == 1) {startX = i;startY = j;} else if (grid[i][j] == 2) {endX = i;endY = j;}}}boolean[][] visited = new boolean[M][N];return process(startX, startY, grid, visited, remain - 1);}// 剩余remain步，從i,j到endX, endY共有多少種路徑public int process(int i, int j, int[][] grid, boolean[][] visited, int remain) {if (i < 0 || i == M || j < 0 || j == N || visited[i][j] || grid[i][j] == -1) return 0;if (remain == 0 || i == endX && j == endY) return remain == 0 && i == endX && j == endY ? 1 : 0;visited[i][j] = true;int p1 = process(i - 1, j, grid, visited, remain - 1);int p2 = process(i + 1, j, grid, visited, remain - 1);int p3 = process(i, j - 1, grid, visited, remain - 1);int p4 = process(i, j + 1, grid, visited, remain - 1);visited[i][j] = false;return p1 + p2 + p3 + p4;} }

總結(jié)

以上是生活随笔為你收集整理的leetcode 62, 63, 980. Unique Paths I, II, III | 62, 63, 980. 不同路径 I, II, III（暴力递归-＞傻缓存-＞动态规划）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

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