LeetCode 64 _ Minimum Path Sum 最小路径和

最新推荐文章于 2022-06-18 22:52:48 发布

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最新推荐文章于 2022-06-18 22:52:48 发布

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原文链接：http://www.cnblogs.com/zingg7/p/10656879.html

本文详细解析了寻找矩阵中从左上角到右下角的最小路径和的问题，并提供了一种有效的动态规划解决方案。通过逐步迭代，算法能够在O(m*n)的时间复杂度内找到最优路径。

Description:

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:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

Solution:

这道题和之前的两题思路差不多的，之前我们在格子里存储的是遍历到该格子的可能情况数，现在则存储遍历到该格子最小的和即可，遍历到该格子的最小和为当前格子加上左边与上面较小的那个值。

Code:

public int minPathSum(int[][] grid) {
    int row = grid.length;
    int col = grid[0].length;
    int[][] res = new int[row][col];


    for (int i = 0; i < row; i++){
        for (int j = 0; j < col; j++){
            if (i == 0 && j == 0){
                res[i][j] = grid[i][j];
            }else if (i == 0){
                res[i][j] = res[i][j-1] + grid[i][j];
            }else if (j == 0){
                res[i][j] = res[i-1][j] + grid[i][j];
            }else{
                res[i][j] = Math.min(res[i-1][j] + grid[i][j], res[i][j-1] + grid[i][j]);
            }
        }
    }
    return res[row-1][col-1];
}

运行情况:

Runtime: 2 ms, faster than 99.04% of Java online submissions for Minimum Path Sum.

Memory Usage: 42.9 MB, less than 13.73% of Java online submissions for Minimum Path Sum.

好像空间有点太大了，也可以不创建新的二维数组，直接存储于原数组，这样空间复杂度会低很多。

转载于:https://www.cnblogs.com/zingg7/p/10656879.html