August 04, 2019

64. Minimum Path Sum

二维 DP Time = O(n^2), Space = O(n^2)

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

下面优化一下 Space Complexity O(n^2) -> O(1)。直接更新当前 2D Array，不需要开辟新的存储空间。

class Solution {
    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        for (int i = 1; i < grid.length; i++) {
            grid[i][0] = grid[i - 1][0] + grid[i][0];
        }
        for (int j = 1; j < grid[0].length; j++) {
            grid[0][j] = grid[0][j - 1] + grid[0][j];
        }
        
        for (int i = 1; i < grid.length; i++) {
            for (int j = 1; j < grid[0].length; j++) {
                grid[i][j] = Math.min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j];
            }
        }
        return grid[grid.length - 1][grid[0].length - 1];
    }
}