【LeetCode】路径系列

62. Unique Paths

题目：从左上到右下，有多少种不同的路径

思路：排列组合

public class Solution {    public int uniquePaths(int m, int n) {        long y = 1, x = 1;        long min = Math.min(m, n);        if(min == 1) return 1;        for(long i = Math.max(m, n); i <= m+n-2; i++){            y *= i;        }        for(long i = 1; i <= min-1; i++){            x *= i;        }        return (int)(y/x);    }}

动态规划：

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

63. Unique Paths II

题目：有1的位置表示有障碍，统计所有路径数

思路：沿用62题动态规划的方法，遇1路径数为零

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

64. Minimum Path Sum

题目：找到最小路径和

思路：动态规划，很简单

public class Solution {    public int minPathSum(int[][] grid) {        int m = grid.length;        int n = grid[0].length;        int[][] dp = new int[m][n];        dp[0][0] = grid[0][0];        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(i ==0 && j > 0) dp[i][j] = dp[i][j-1]+grid[i][j];                if(j ==0 && i > 0) dp[i][j] = dp[i-1][j]+grid[i][j];                if(i > 0 && j > 0) dp[i][j] = Math.min(dp[i][j-1], dp[i-1][j]) + grid[i][j];            }        }        return dp[m-1][n-1];    }}

70. Climbing Stairs

题目：爬梯子，每步1到2级，n步能爬到多少级

思路：斐波那契数列，没啥好说的

public class Solution {    public int climbStairs(int n) {        if(n < 2) return 1;        int pre = 1;        int x = 1;        for(int i = 2; i <= n; i++){            int next = pre+x;            pre = x;            x = next;        }        return x;    }}