Difficulty:: Medium
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:
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| 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
Language: Java
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| class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
for (int i = 1; i < n; i++) {
grid[0][i] += grid[0][i - 1];
}
for (int i = 1; i < m; i++) {
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 - 1][j], grid[i][j - 1]);
}
}
return grid[m - 1][n - 1];
}
}
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| class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[2][n];
dp[0][0] = grid[0][0];
for (int i = 1; i < n; i++) {
dp[0][i] = dp[0][i - 1] + grid[0][i];
}
for (int i = 1; i < m; i++) {
for (int j = 0; j < n; j++) {
if (j == 0) {
dp[i % 2][j] = dp[(i - 1) % 2][j] + grid[i][j];
} else {
dp[i % 2][j] = Math.min(dp[(i - 1) % 2][j], dp[i % 2][j - 1]) + grid[i][j];
}
}
}
return dp[(m - 1) % 2][n - 1];
}
}
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