[刷题]Minimum Path Sum

最新推荐文章于 2024-11-20 16:53:47 发布

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本文介绍了一个LintCode上的题目“最小路径和”的解决方案。该题要求在一个二维网格中找到从左上角到右下角路径上的最小和。通过动态规划的方法，先初始化边界条件，再逐步填充内部单元格，最终返回右下角单元格的值作为答案。

[LintCode]Minimum Path Sum

public class Solution {
    /**
     * @param grid: a list of lists of integers.
     * @return: An integer, minimizes the sum of all numbers along its path
     */
    public int minPathSum(int[][] grid) {
        // 2015-05-13
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        
        int m = grid.length;
        int n = grid[0].length;
        int[][] sum = new int[m][n];
        
        sum[0][0] = grid[0][0];
        for (int i = 1; i < m; i++) {
            sum[i][0] = sum[i - 1][0] + grid[i][0];
        }
        for (int j = 1; j < n; j++) {
            sum[0][j] = sum[0][j - 1] + grid[0][j];
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                sum[i][j] = Math.min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
            }
        }
        return sum[m - 1][n - 1];
    }
}