Minimum Path Sum問題

該問題來自 LeetCode 的第64題。
64. Minimum Path Sum

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.

Example 1:
[[1,3,1],
[1,5,1],
[4,2,1]]
Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.

問題分析

該問題和另外一篇文章《 JAVA程式碼—演算法基礎：矩陣的最小路徑和問題》描述的其實是一類問題。連結：http://blog.csdn.net/seagal890/article/details/79590804 。都可以採用動態規劃（Dynamic Programming）演算法求解。

public int minPathSum(int[][] grid) {
    int m = grid.length;// row
    int n = grid[0].length; // column
    for (int i = 0; i < m; i++) {
        for (int j = 0
 
; j < n; j++) {
            if (i == 0 && j != 0) {
                grid[i][j] = grid[i][j] + grid[i][j - 1];
            } else if (i != 0 && j == 0) {
                grid[i][j] = grid[i][j] + grid[i - 1][j];
            } else if (i == 0 && j == 0) {
                grid[i][j] = grid
 
[i][j];
            } else {
                grid[i][j] = Math.min(grid[i][j - 1], grid[i - 1][j])
                        + grid[i][j];
            }
        }
    }

    return grid[m - 1][n - 1];
}

簡化一下演算法：

public int minPathSum(int[][] grid) {
    int m = grid.length;// row
    int n = grid[0].length; // column
    for(int j = 1; j < n; j++) grid[0][j] = grid[0][j] + grid[0][j-1];
    for(int i = 1; i < m; i++) grid[i][0] = grid[i][0] + grid[i-1][0];
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
                grid[i][j] = Math.min(grid[i][j - 1], grid[i - 1][j]) + grid[i][j];
        }
    }
    return grid[m - 1][n - 1];
}

（完）