62. Unique Paths
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题目:
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.
题意:
如上图所示,左上的机器人到右下的星星处有多少种不同的走法。
思路:
DP问题。动态规划问题。题意可知一个格子到另一个格子只能向下或者向右。故第一行,第一列到达的唯一路径都只有一种可能,则result[0][j]= result[i][0]=1。而其他格子到达的路径result[i][j] = reesult[i-1][j] + result[i][j-1];
代码:class Solution {public: int uniquePaths(int m, int n) { vector<vector<int>> result(m, vector<int>(n, 1)); for(int i=1; i<m; i++){ for(int j=1; j<n; j++){ result[i][j] = result[i-1][j] + result[i][j-1]; } } return result[m-1][n-1]; }};上面代码中保存了每个格子对应到达的唯一路径数,题目只要求得到唯一的整数,不需要保存到达每个格子的唯一路径数,故只需要维持两个数组保存向下与向右方向的唯一路径数即可。
代码:
class Solution {public: int uniquePaths(int m, int n) { if(m>n){ return uniquePaths(n, m); } vector<int> pre(m, 1); vector<int> cur(m, 1); for(int i=1; i<n; i++){ for(int j=1; j<m; j++){ cur[j] = cur[j-1]+pre[j]; } swap(pre, cur); } return pre[m-1]; }};还可以进一步缩紧所花费的空间复杂度。不要维持一个pre数组。直接在元素上进行修改。
代码:
class Solution {public: int uniquePaths(int m, int n) { if(m>n){ return uniquePaths(n, m); } vector<int> cur(m, 1); for(int i=1; i<n; i++){ for(int j=1; j<m; j++){ cur[j] += cur[j-1]; } } return cur[m-1]; }};转载地址:https://leetcode.com/discuss/38353/0ms-5-lines-dp-solution-in-c-with-explanations
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