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:7Explanation:Because the path 1→3→1→1→1 minimizes the sum.
class Solution {
public int minPathSum(int[][] grid) {
if (grid.length == 0) return 0;
int sum[][] = new int[grid.length][grid[0].length];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (i != 0 && j != 0) sum[i][j] = grid[i][j] + Math.min(sum[i - 1][j], sum[i][j - 1]);
else if (i == 0 && j == 0) sum[i][j] = grid[i][j];
else if (i == 0) sum[i][j] = grid[i][j] + sum[i][j - 1];
else if (j == 0) sum[i][j] = grid[i][j] + sum[i - 1][j];
}
}
// System.out.println(Arrays.deepToString(sum));
return sum[grid.length - 1][grid[0].length - 1];
}
}
(这题看了答案，我的答案排名83%，靠前的跟我看起来差不多，也不知道是哪里有不同)
第二种解法：
与第一种类似，但是只记录一行的dp就行，空间复杂度减少，但是个人感觉这个代码稍微难理解一点：
class Solution {
public int minPathSum(int[][] grid) {
if (grid.length == 0) return 0;
int[] dp = new int[grid[0].length];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (j == 0) {
dp[j] = dp[j] + grid[i][j];
} else if (i == 0) {
dp[j] = dp[j - 1] + grid[i][j];
} else {
dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j];
}
}
}
return dp[grid[0].length - 1];
}
}
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