LeetCode62 Unique Paths

動態規劃
dp[i][j] = dp[i-1][j]+dp[i][j-1]

class Solution {
    public int uniquePaths(int m, int n) {
        if(m == 0 || n == 0) return 0;
        if(m == 1 || n == 1) return 1;
        int[][] dp = new int[m][n];
        for(int i = 0 ; i < m ; i++) dp[i][0] = 1;
        for(int i = 0 ; i < n ; i++) dp[0][i] = 1;
        dp[0][0] = 0;
        for(int i = 1 ; i < m ; i++) {
            for(int j = 1 ; j < n ; j++) {
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
}

空間優化：
我看的一位博主寫的很好，鏈接放這裏Leetcode62

class Solution {
    public int uniquePaths(int m, int n) {
        if(m == 0 || n == 0) return 0;
        if(m == 1 || n == 1) return 1;
        int[] dp = new int[n];
        dp[0] = 1;
        for(int i = 0 ; i < m ; i++) {
            for(int j = 1 ; j < n ; j++) {
                dp[j] +=  dp[j-1];
            }
        }
        return dp[n-1];
    }
}

這道題不同於62的區別是有障礙物
時間複雜度：O(m*n)
空間複雜度：O(1)

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        if(obstacleGrid[0][0] == 1 || obstacleGrid[m-1][n-1] == 1) return 0;
        
        obstacleGrid[0][0] = 1;
        for(int i = 1 ;  i < m ; i++) obstacleGrid[i][0] = (obstacleGrid[i-1][0] == 0 || obstacleGrid[i][0] == 1) ? 0 : 1;
        for(int i = 1 ; i < n ; i++) obstacleGrid[0][i] = (obstacleGrid[0][i-1] == 0 || obstacleGrid[0][i] == 1) ? 0 : 1;
        for(int i = 1 ; i < m ; i++) {
            for(int j = 1 ; j < n ; j++) {
                if(obstacleGrid[i][j] == 1) obstacleGrid[i][j] = 0;
                else obstacleGrid[i][j] = obstacleGrid[i-1][j] + obstacleGrid[i][j-1];
            }
        }
        return obstacleGrid[m-1][n-1];
    }
}