【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

Unique Paths（系列）

Unique Paths A robot is located at the top-left corner of a   m   x   n   grid (marked 'Start' in the diagram below). The robot can only move either down or right a t any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). How many possible unique paths are there?

 

Above is a 7 x 3 grid. How many possible unique paths are there? Note:   m   and   n   will be at most 100. Example 1: Input: m = 3, n = 2 Output: 3 Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 1. Right -> Right -> Down 2. Right -> Down -> Right 3. Down -> Right -> Right Example 2: Input: m = 7, n = 3 Output: 28

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C++   //問題：街區路徑問題(與LCS最長公共子序列問題類似) 求機器人從起點到終點的路徑數量 //方法一：設P[i][j] 為到座標（i,j）的路徑總數，則由於機器人 只能向右或者向下 ，故有P[i][j] = P[i - 1][j] + P[i][j - 1] //O(n^2) O(m*n) class Solution { public :     int uniquePaths ( int m , int n )     {         vector < vector < int >> path ( m , vector < int >( n , 1 )); //(1,0)和（0,1）處初始化為1                for ( int i = 1 ; i < m ; i ++) //i從1開始，掃描行             for ( int j = 1 ; j < n ; j ++) //掃描列                 path [i][j] = path[i-1][j] + path[i][j-1];                return path [ m - 1 ][ n - 1 ];     } }; /* * 方法一改進   用一個列向量存當前需要的元素即可，將空間複雜度優化至O（min(m,n)） * 例子     1 1 1 1  1  1     1 2 3 4  5  6     1 3 6 10 15 21 dp[i][j] = dp[i - 1][j] + dp[i][j - 1] dp[i][j],dp[i-1][j] ---> dp[j] dp[i][j-1] ---> dp[j-1] */ class Solution { public :     int uniquePaths ( int m , int n )     {         vector < int > dp ( n , 1 ); //進一步改進可以選擇min(m,n)                 for ( int i = 1 ; i < m ; ++ i ) //掃描行         {             for ( int j = 1 ; j < n ; ++ j ) //掃描列             {                 dp [ j ] += dp [ j - 1 ];             }         }         return dp [ n - 1 ];     } };   63 .   Unique Paths II Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as   1   and   0   respectively in the grid. Note:   m   and   n   will be at most 100. Example 1: Input: [ [0,0,0], [0,1,0], [0,0,0] ] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right

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  /* 問題：有障礙物的柵格中，機器人從起始點到終點的路徑數量 方法：動態規劃 對於有障礙物的柵格中，dp值置0 */ class Solution { public :     int uniquePathsWithObstacles ( vector < vector < int >>& obstacleGrid )     {         if ( obstacleGrid . empty () || obstacleGrid [ 0 ]. empty () || obstacleGrid [ 0 ][ 0 ] == 1 )             return 0 ; //矩陣為空或者起始點為障礙點時，退出                 int m = obstacleGrid . size ();         int n = obstacleGrid [ 0 ]. size ();         vector < vector < int >> dp ( m , vector < int >( n )); //初始化為0                 for ( int i = 0 ; i < m ; i ++) //從0開始掃描，因為任何點都有可能為障礙點(如（0,1）和（1,0）)         {             for ( int j = 0 ; j < n ; j ++)             {                 if ( obstacleGrid [ i ][ j ] == 1 ) dp [ i ][ j ] = 0 ; //有障礙的地方dp值置0                 else if ( i == 0 && j == 0 ) dp [ i ][ j ] = 1 ;    //起始點到起始點路徑數為1（若無障礙）                 else if ( i == 0 && j > 0 ) dp [ i ][ j ] = dp [ i ][ j - 1 ]; //第一行的處理                 else if ( i > 0 && j == 0 ) dp [ i ][ j ] = dp [ i - 1 ][ j ]; //第一列的處理                 else dp [ i ][ j ] = dp [ i - 1 ][ j ] + dp [ i ][ j - 1 ]; //其他情況的處理             }         }         return dp . back (). back (); //返回末尾點     } };