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1.
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.
class Solution { public: int uniquePaths(int m, int n) { vector<vector<int>> v(m, vector<int>(n, 0)); int i, j; for(i = 0; i < m; i++) { for(j = 0; j < n; j++) { if(0 == i || 0 == j) v[i][j] = 1; else v[i][j] = v[i-1][j] + v[i][j-1]; } } return v[m-1][n-1]; } };
2.
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.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[ [0,0,0], [0,1,0], [0,0,0] ]
The total number of unique paths is 2.
Note: m and n will be at most 100.
class Solution { public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int m = obstacleGrid.size(); if(m <= 0) return 0; int n = obstacleGrid[0].size(); if(n <= 0 || obstacleGrid[0][0] == 1) return 0; vector<vector<int>> v(m, vector<int>(n, 0)); int i, j; for(i = 0; i < m; i++) { for(j = 0; j < n; j++) { if(obstacleGrid[i][j] == 1) v[i][j] = 0; else if(0 == i && j) v[i][j] = v[0][j-1]; else if(0 == j && i) v[i][j] = v[i-1][0]; else if(i && j) v[i][j] = v[i-1][j] + v[i][j-1]; else v[i][j] = 1; } } return v[m-1][n-1]; } };
3.
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
int minPathSum(vector<vector<int> > &grid) { if (grid.size()<=0){ return 0; } int i, j; for(i=0; i<grid.size(); i++){ for(j=0; j<grid[i].size(); j++){ int top = i-1<0 ? INT_MAX : grid[i-1][j] ; int left = j-1<0 ? INT_MAX : grid[i][j-1]; if (top==INT_MAX && left==INT_MAX){ continue; } grid[i][j] += (top < left? top: left); } } return grid[grid.size()-1][grid[0].size()-1]; }
62. 63. Unique Paths 64. Minimum Path Sum
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原文地址：http://www.cnblogs.com/argenbarbie/p/5270183.html
