64. Minimum Path Sum
摘要：
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 pat...
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.
難度：medium
題目：給定m * n的格子，找出從左上角到右下角和最小的路徑。
思路：動態規劃，狀態轉移方程
grid[i][j] = grid[i][j] + min(grid[i - 1][j], grid[i][j - 1]
Runtime: 4 ms, faster than 97.83% of Java online submissions for Minimum Path Sum.
Memory Usage: 40.8 MB, less than 1.47% of Java online submissions for Minimum Path Sum.
class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length, 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];
}
}