[LintCode] Minimum Path Sum

时间：2017-11-12 12:23:38 阅读：135 评论：0 收藏：0 [点我收藏+]

标签：tree rom tom += mic code void pat can

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.

You can only move either down or right at any point in time.

Solution 1. Recursion, DFS

 1 public class Solution {
 2     private int minPath = Integer.MAX_VALUE;
 3     public int minPathSum(int[][] grid) {
 4         if(grid == null || grid.length == 0 || grid[0].length == 0){
 5             return -1;
 6         }
 7         traverse(grid, 0, 0, 0);
 8         return minPath;
 9     }
10     private void traverse(int[][] grid, int x, int y, int sum){
11         int m = grid.length, n = grid[0].length;
12         if(x >= m || y >= n) {
13             return;
14         }
15         sum += grid[x][y];
16         if(x == m - 1 && y == n - 1){
17             if(sum < minPath){
18                 minPath = sum;
19             }
20             return;
21         }
22         traverse(grid, x + 1, y, sum);  
23         traverse(grid, x, y + 1, sum); 
24     }
25 }

Solution 2. DFS, divide and conquer + memoization

 1 public class Solution {
 2     private int[][] hash;
 3     public int minPathSum(int[][] grid) {
 4         if(grid == null || grid.length == 0 || grid[0].length == 0){
 5             return -1;
 6         }
 7         hash = new int[grid.length][grid[0].length];
 8         for(int i = 0; i < hash.length; i++){
 9             for(int j = 0; j < hash[0].length; j++){
10                 hash[i][j] = Integer.MAX_VALUE;
11             }
12         }
13         return divideConquer(grid, 0, 0);
14     }
15     private int divideConquer(int[][] grid, int x, int y){
16         int m = grid.length, n = grid[0].length;
17         if(x == m - 1 && y == n - 1){
18             return grid[x][y];
19         }
20         if(x >= m || y >= n){
21             return Integer.MAX_VALUE;
22         }
23         if(hash[x][y] != Integer.MAX_VALUE){
24             return hash[x][y];
25         }
26         hash[x][y] = grid[x][y] + Math.min(divideConquer(grid, x, y + 1), divideConquer(grid, x + 1, y));
27         return hash[x][y];
28     }
29 }

Solution 3. Dynamic Programming

State: f[i][j], the min distance from (0, 0) to (i, j)

Function: f[i][j] = grid[i][j] + Math.min(f[i - 1][j], f[i][j - 1])

Initialization: f[i][0] = grid[i][0] + f[i - 1][0]; f[0][j] = grid[0][j] + f[0][j - 1]

Answer: f[m - 1][n - 1]

 1 public class Solution {
 2     public int minPathSum(int[][] grid) {
 3         if(grid == null || grid.length == 0 || grid[0].length == 0){
 4             return -1;
 5         }
 6         int rowLen = grid.length, colLen = grid[0].length;
 7         int[][] f = new int[rowLen][colLen];
 8         f[0][0] = grid[0][0];
 9         for(int row = 1; row < rowLen; row++){
10             f[row][0] = grid[row][0] + f[row - 1][0]; 
11         }
12         for(int col = 1; col < colLen; col++){
13             f[0][col] = grid[0][col] + f[0][col - 1];
14         }
15         for(int i = 1; i < rowLen; i++){
16             for(int j = 1; j < colLen; j++){
17                 f[i][j] = grid[i][j] + Math.min(f[i][j - 1], f[i - 1][j]);
18             }
19         }
20         return f[rowLen - 1][colLen - 1];
21     }
22 }

Solution 4. DP with space optimization

 1 public class Solution {
 2     public int minPathSum(int[][] grid) {
 3         if(grid == null || grid.length == 0 || grid[0].length == 0){
 4             return -1;
 5         }
 6         int rowLen = grid.length, colLen = grid[0].length;
 7         int[][] T = new int[2][colLen];
 8         T[0][0] = grid[0][0];
 9         for(int j = 1; j < colLen; j++){
10             T[0][j] = T[0][j - 1] + grid[0][j];
11         }
12         for(int i = 1; i < rowLen; i++){
13             T[i % 2][0] = T[(i - 1) % 2][0] + grid[i][0];
14             for(int j = 1; j < colLen; j++){
15                 T[i % 2][j] = grid[i][j] + Math.min(T[(i - 1) % 2][j], T[i % 2][j - 1]);
16             }
17         }
18         return T[(rowLen - 1) % 2][colLen - 1];
19     }
20 }