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minimun-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:
Input:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.
This is a simple DP problem, current sum can either come from up or left, since it can only move down or right.
For each sum,
grid[r][c] = min(grid[r-1][c], grid[r][c-1]) + grid[r][c]; then you find the min path to current position (r,c).after scan the whole grid, return bottom right value.
NOTE: Below implementation, modified original input grid, from algorithm point of view to reduce space complexity to
O(1). But from real project, modified original input not recommended, use extra space.For example:
Minimum Path Sum
Time Complexity:
O(N*M)Space Complexity:
O(1)- N - number of grid rows
- M - number of grid columns
class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
int row = grid.length;
int col = grid[0].length;
// init first row, since only move down or right, first line can only move from left to right
for (int c = 1; c < col; c++) {
grid[0][c] += grid[0][c - 1];
}
// init first col, only move down
for (int r = 1; r < row; r++) {
grid[r][0] += grid[r - 1][0];
}
for (int r = 1; r < row; r++) {
for (int c = 1; c < col; c++) {
grid[r][c] += Math.min(grid[r - 1][c], grid[r][c - 1]);
}
}
return grid[row - 1][col - 1];
}
}
Last modified 3yr ago