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.
Code from http://www.darrensunny.me/leetcode-minimum-path-sum/
// DP, O(n^2) time, O(n^2) space
Code from http://joycelearning.blogspot.com/2013/10/leetcode-minimum-path-sum.html
Read full article from 一天一学: Leetcode - Minimum Path Sum
Note: You can only move either down or right at any point in time.
Code from http://www.darrensunny.me/leetcode-minimum-path-sum/
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0)
return 0;
int m = grid.length, n = grid[0].length;
int[] dp = new int[n];
// Find the minimum sum of all numbers along a path to grid[i][j]
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j == 0) // The first cell
dp[j] = grid[i][j];
else if (i == 0) // Cells in the first row
dp[j] = dp[j-1] + grid[i][j];
else if (j == 0) // Cells in the first column
dp[j] += grid[i][j];
else // Others
dp[j] = Math.min(dp[j-1], dp[j]) + grid[i][j];
}
}
return dp[n-1];
}
// DP, O(n^2) time, O(n^2) space
public
int
minPathSum(
int
[][] grid) {
int
row = grid.length;
int
col = grid[
0
].length;
int
[][] res =
new
int
[row][col];
// init
res[
0
][
0
] = grid[
0
][
0
];
// left
for
(
int
i =
1
; i < row; i++) {
res[i][
0
] = res[i -
1
][
0
] + grid[i][
0
];
}
// top
for
(
int
j =
1
; j < col; j++) {
res[
0
][j] = res[
0
][j -
1
] + grid[
0
][j];
}
// rest elements
for
(
int
i =
1
; i < row; i++) {
for
(
int
j =
1
; j < col; j++) {
res[i][j] = grid[i][j] + Math.min(res[i -
1
][j], res[i][j -
1
]);
}
}
return
res[row -
1
][col -
1
];
}
public
int
minPathSum(
int
[][] grid) {
int
row = grid.length;
int
col = grid[
0
].length;
int
[] res =
new
int
[col];
// init
Arrays.fill(res, Integer.MAX_VALUE);
res[
0
] =
0
;
// rest elements
for
(
int
i =
0
; i < row; i++) {
// init the 0th sum = old 0th element + the new 0th element
// just init the 0th column every time dynamically
res[
0
] = res[
0
] + grid[i][
0
];
// loop through each element of each row
for
(
int
j =
1
; j < col; j++) {
res[j] = grid[i][j] + Math.min(res[j], res[j -
1
]);
}
}
return
res[col -
1
];
}
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