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[LeetCode] 764. Largest Plus Sign #764

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grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 764. Largest Plus Sign #764

grandyang opened this issue May 30, 2019 · 0 comments

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@grandyang
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grandyang commented May 30, 2019

 

In a 2D grid from (0, 0) to (N-1, N-1), every cell contains a 1, except those cells in the given list mines which are 0. What is the largest axis-aligned plus sign of 1s contained in the grid? Return the order of the plus sign. If there is none, return 0.

An " axis-aligned plus sign of1s  of order k" has some center grid[x][y] = 1 along with 4 arms of length k-1going up, down, left, and right, and made of 1s. This is demonstrated in the diagrams below. Note that there could be 0s or 1s 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
010
000

Order 2:
00000
00100
01110
00100
00000

Order 3:
0000000
0001000
0001000
0111110
0001000
0001000
0000000

 

Example 1:

Input: N = 5, mines = [[4, 2]]
Output: 2
Explanation:
11111
11111
11111
11111
11011
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:

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

 

这道题给了我们一个数字N,表示一个NxN的二位数字,初始化均为1,又给了一个mines数组,里面是一些坐标,表示数组中这些位置都为0,然后让我们找最大的加型符号。所谓的加型符号是有数字1组成的一个十字型的加号,题目中也给出了长度分别为1,2,3的加型符号的样子。好,理解了题意以后,我们来想想该如何破题。首先,最简单的就是考虑暴力搜索啦,以每个1为中心,向四个方向分别去找,只要任何一个方向遇到了0就停止,然后更新结果res。令博主感到惊讶的是,此题的OJ居然允许Brute Force的解法通过,还是比较大度的,参见代码如下:

 

解法一:

class Solution {
public:
    int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) {
        int res = 0;
        vector<vector<int>> mat(N, vector<int>(N, 1));
        for (auto mine : mines) mat[mine[0]][mine[1]] = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                int k = 0;
                while (canExpand(mat, N, i, j, k)) ++k;
                res = max(res, k);
            }
        }
        return res;
    }
    bool canExpand(vector<vector<int>>& mat, int N, int x, int y, int k) {
        if (x - k < 0 || y - k < 0 || x + k >= N || y + k >= N) return false;
        return mat[x - k][y] && mat[x][y + k] && mat[x + k][y] && mat[x][y - k];
    }
};

 

如果我们只想出暴力搜索的解法,就不再管这道题了的话,那在面试的时候就比较悬了。毕竟立方级的时间复杂度实在是太高了,我们必须要进行优化。暴力搜索的时间复杂度之所以高的原因是因为对于每一个1都要遍历其上下左右四个方向,有大量的重复计算,我们为了提高效率,可以对于每一个点,都计算好其上下左右连续1的个数。博主最先用的方法是建立四个方向的dp数组,dp[i][j]表示 (i, j) 位置上该特定方向连续1的个数,那么就需要4个二维dp数组,举个栗子,比如:

原数组:

1  0  1  0
1  1  1  1
1  0  1  1

那么我们建立left数组是当前及其左边连续1的个数,如下所示:

1  0  1  0
1  2  3  4
1  0  1  2

right数组是当前及其右边连续1的个数,如下所示:

1  0  1  0
4  3  2  1
1  0  2  1

up数组是当前及其上边连续1的个数,如下所示:

1  0  1  0
2  1  2  1
3  0  3  2

down数组是当前及其下边连续1的个数,如下所示:

3  0  3  0
2  1  2  2
1  0  1  1

我们需要做的是在这四个dp数组中的相同位置的四个值中取最小的一个,然后在所有的这些去除的最小值中选最大一个返回即可。为了节省空间,我们不用四个二维dp数组,而只用一个就可以了,因为对于每一个特定位置,我们只需要保留较小值,所以在更新的时候,只需要跟原来值相比取较小值即可。在计算down数组的时候,我们就可以直接更新结果res了,因为四个值都已经计算过了,我们就不用再重新在外面开for循环了,参见代码如下:

 

解法二:

class Solution {
public:
    int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) {
        int res = 0, cnt = 0;
        vector<vector<int>> dp(N, vector<int>(N, 0));
        unordered_set<int> s;
        for (auto mine : mines) s.insert(mine[0] * N + mine[1]);
        for (int j = 0; j < N; ++j) {
            cnt = 0;
            for (int i = 0; i < N; ++i) { // up
                cnt = s.count(i * N + j) ? 0 : cnt + 1;
                dp[i][j] = cnt;
            }
            cnt = 0;
            for (int i = N - 1; i >= 0; --i) { // down
                cnt = s.count(i * N + j) ? 0 : cnt + 1;
                dp[i][j] = min(dp[i][j], cnt);
            }
        }
        for (int i = 0; i < N; ++i) {
            cnt = 0;
            for (int j = 0; j < N; ++j) { // left
                cnt = s.count(i * N + j) ? 0 : cnt + 1;
                dp[i][j] = min(dp[i][j], cnt);
            }
            cnt = 0;
            for (int j = N - 1; j >= 0; --j) { // right
                cnt = s.count(i * N + j) ? 0 : cnt + 1;
                dp[i][j] = min(dp[i][j], cnt);
                res = max(res, dp[i][j]);
            }
        }
        return res;
    }
};

 

我们可以进一步的压缩代码,使其更加简洁,我们发现其实只要分别用四个变量l,r,u,d来表示四个方向连续1的个数,既可以将for循环糅合在一起。注意里面内嵌的for循环其实是两个for循环,由j和k分别控制,那么只要弄清i,j,k坐标的位置,就可以同时更新四个方向的dp值了,最后dp数组更新好了之后,我们再秀一波,只用一个for循环来遍历二维数组,其实就是把二维坐标压缩成了一个数字,再解压缩,参见代码如下:

 

解法三:

class Solution {
public:
    int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) {
        int res = 0;
        vector<vector<int>> dp(N, vector<int>(N, N));
        for (auto mine : mines) dp[mine[0]][mine[1]] = 0;
        for (int i = 0; i < N; ++i) {
            int l = 0, r = 0, u = 0, d = 0;
            for (int j = 0, k = N - 1; j < N; ++j, --k) {
                dp[i][j] = min(dp[i][j], l = (dp[i][j] ? l + 1 : 0));
                dp[j][i] = min(dp[j][i], u = (dp[j][i] ? u + 1 : 0));
                dp[i][k] = min(dp[i][k], r = (dp[i][k] ? r + 1 : 0));
                dp[k][i] = min(dp[k][i], d = (dp[k][i] ? d + 1 : 0));
            }
        }
        for (int k = 0; k < N * N; ++k) res = max(res, dp[k / N][k % N]);
        return res;
    }
};

 

类似题目:

Cheapest Flights Within K Stops

Minimum Swaps To Make Sequences Increasing

Soup Servings

 

参考资料:

https://leetcode.com/problems/largest-plus-sign/solution/

https://leetcode.com/problems/largest-plus-sign/discuss/113314/JavaC++Python-O(N2)-solution-using-only-one-grid-matrix

https://leetcode.com/problems/largest-plus-sign/discuss/113350/C++-simple-brute-force-easy-to-understand-with-detailed-explanation

 

LeetCode All in One 题目讲解汇总(持续更新中...)

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