Largest Plus Sign(DP)

In a 2Dgridfrom (0, 0) to (N-1, N-1), every cell contains a1, except those cells in the given listmineswhich are0. What is the largest axis-aligned plus sign of1s contained in the grid? Return the order of the plus sign. If there is none, return 0.

An "axis-aligned plus sign of1sof orderk" has some centergrid[x][y] = 1along with 4 arms of lengthk-1going up, down, left, and right, and made of1s. This is demonstrated in the diagrams below. Note that there could be0s or1s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1s.

Examples of Axis-Aligned Plus Signs of Order k:

Order 1:
000
0
1
0
000

Order 2:
00000
00
1
00
0
111
0
00
1
00
00000

Order 3:
0000000
000
1
000
000
1
000
0
11111
0
000
1
000
000
1
000
0000000

Example 1:

Input:
 N = 5, mines = [[4, 2]]

Output:
 2

Explanation:

11111
11111
1
1
111

111
11
1
1
011
In the above grid, the largest plus sign can only be order 2.  One of them is marked in bold.

Example 2:

Input:
 N = 2, mines = []

Output:
 1

Explanation:

There is no plus sign of order 2, but there is of order 1.

Example 3:

Input:
 N = 1, mines = [[0, 0]]

Output:
 0

Explanation:

There is no plus sign, so return 0.

Note:

N will be an integer in the range [1, 500].
mines will have length at most 5000.
mines[i] will be length 2 and consist of integers in the range [0, N-1].
(Additionally, programs submitted in C, C++, or C# will be judged with a slightly smaller time limit.)

分析

dp里存该点开始能延伸到的最大的边长，遇到0就重置为0。4个方向边长都存在这个dp点上，取4个方向里最小的边长。

最后结果就是dp里最大的边长。

1.建一个int[][]dp，fill N,在mines里为0，。

2.4个方向尽量延伸得到l,r,u,d，遇到0cell就重置为0。dp都取最小值，这个每个dp里都是存的4个方向里最小的边长。

3.最后整个dp里最大的那个就是最大边长。

class Solution {
    public int orderOfLargestPlusSign(int N, int[][] mines) {
        int[][] dp = new int[N][N];
        for (int i = 0; i < N; i++) {
            Arrays.fill(dp[i], N);
        }
        for(int[] p : mines){
            dp[p[0]][p[1]] = 0;
        }

        for (int i = 0; i < N; i++){
//             for(int j = 0, l = 0; j < N; j ++){
//                 dp[i][j] = Math.min(dp[i][j], l = dp[i][j] == 0 ? 0 : l + 1);
//             }

//             for(int j = N-1, r = 0; j >= 0; j --){
//                 dp[i][j] = Math.min(dp[i][j], r = dp[i][j] == 0 ? 0 : r + 1);
//             }

//             for(int j = 0, u = 0; j < N; j ++){
//                 dp[j][i] = Math.min(dp[j][i], u = dp[j][i] == 0 ? 0 : u + 1);
//             }

//             for(int j = N-1, d = 0; j >= 0; j --){
//                 dp[j][i] = Math.min(dp[j][i], d = dp[j][i] == 0 ? 0 : d + 1);
//             }

            //注意这种简写法。
            for(int j=0,l=0,r=0,u=0,d = 0, k = N-1; j < N; j ++, k --){
                dp[i][j] = Math.min(dp[i][j], l = dp[i][j] == 0 ? 0 : l + 1);

                dp[i][k] = Math.min(dp[i][k], r = dp[i][k] == 0 ? 0 : r + 1);

                dp[j][i] = Math.min(dp[j][i], u = dp[j][i] == 0 ? 0 : u + 1);

                dp[k][i] = Math.min(dp[k][i], d = dp[k][i] == 0 ? 0 : d + 1);
            }

        }

        int o = 0;
        for (int i = 0; i < N; i++) {
            for (int j = 0, l = 0; j < N; j++) {
                o = Math.max(dp[i][j],o);
            }
        }
        return o;
    }
}