題目：

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 at 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 3 x 7 grid. How many possible unique paths are there?

Note:?m?and?n?will be at most 100.

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思路1：遞歸

終點的路徑數= 所有能達到終點的點的路徑數之和，即為：路徑[m][n] =?路徑[m-1][n] +?路徑[m][n-1]

依次遞歸，直到start點（0，0）

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代碼：

public static int uniquePaths(int m, int n) {if(n == 1 && m==1){//初始的位置return 1;}else if(n == 1 && m>1){return uniquePaths(1,m-1);}else if(n >1 && m==1){return uniquePaths(n-1,1);}else{return uniquePaths(n-1,m)+uniquePaths(n,m-1);}}

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結果在m=17 n=23的時候超時了。所以意識到題目的限制條件是需要速度！

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思路2：動態規劃

某一個點的路徑數= 所有能達到該點的點的路徑數之和，即為：路徑[m][n] =?路徑[m-1][n] +?路徑[m][n-1]

與遞歸不同的是，遞歸是從大往小算，動態規劃是從小往大算

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代碼如下：

public static int uniquePaths(int m, int n) {int[][] arrResult = new int[m][n];for(int i = 0;i<m;i++){for(int j = 0;j<n;j++){if(i == 0 && j== 0){ arrResult[0][0] = 1;}else if(i == 1 && j== 0){arrResult[1][0] = 1;}else if(i == 0 && j== 1){arrResult[0][1] = 1;}//以上是填充基礎值else{int row = 0;int column = 0;if(i> 0 ){row = arrResult[i-1][j];}if(j> 0 ){column = arrResult[i][j-1];}arrResult[i][j] = row + column; }}}return arrResult[m-1][n-1];}

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轉載于:https://www.cnblogs.com/savageclc26/p/4871140.html

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