【leetcode】Largest Plus Sign

题目如下：

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 of 1s of order k" has some center grid[x][y] = 1 along with 4 arms of length k-1 going 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:

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.)

 

解题思路如下：

首先把二维数组构造出来，然后把mines所在的位置标记为0，非mines标记为1，这是基本。接下来遍历所有的非mines，计算其上下左右四个方向最多相邻值为1的个数，然后取四个值的最小值，即为该位置的Plus Sign，最后求出所有Plus Sign的最大值即可。这里有一点可以优化的地方，就是计算位置Plus Sign的时候，可以每次计算出一个方向后与之前已经计算出的位置的Largest值比较，如果小于Largest，其他方向就可以不用计算了，直接continue到下一个位置。

 

完整代码如下：

/**
 * @param {number} N
 * @param {number[][]} mines
 * @return {number}
 */
largest = 0
var calcUp = function(M,N,x,y){
    var count = 0
    var c = x
    while(--c >= 0){
        if(M[c][y] == 1){
            count++
        }
        else{
            break
        }
    }
    return count
}

var calcDown = function(M,N,x,y){
    var count = 0
    var c = x
    while(++c < N ){
        if(M[c][y] == 1){
            count++
        }
        else{
            break
        }
    }
    return count
}

var calcLeft = function(M,N,x,y){
    var count = 0
    var c = y
    while(--c >=0){
        if(M[x][c] == 1){
            count++
        }
        else{
            break
        }
    }
    return count
}

var calcRight = function(M,N,x,y){
    var count = 0
    var c = y
    while(++c < N){
        if(M[x][c] == 1){
            count++
        }
        else{
            break
        }
    }
    return count
}


var calcLargest = function(M,N,x,y) {
    var up = calcUp(M,N,x,y)
    if (up < largest){
        return -1
    }
    var down = calcDown(M,N,x,y)
    if (down < largest ){
        return -1
    }
    var left = calcLeft(M,N,x,y)
    if (left < largest){
        return -1
    }
    var right = calcRight(M,N,x,y)
    if (right < largest){
        return -1
    }
    return Math.min(up,down,left,right)
}
var orderOfLargestPlusSign = function(N, mines) {
    largest = 0
    var mar = []
    for(var i = 0;i<N;i++){
        var row = []
        for(var j = 0;j<N;j++){
            row.push(1)
        }
        mar.push(row)
    }
    for(var i=0;i<mines.length;i++){
        mar[mines[i][0]][mines[i][1]] = 0
    }
    for(var i = 0;i<N;i++){
        for(var j = 0;j<N;j++){
            if(mar[i][j] == 0){
                continue
            }
            else{
                var r = calcLargest(mar,N,i,j)
                //console.log(i,j,r)
                if (r == -1){
                    continue
                }
                var l = 1 + r
                if (largest < l){
                    largest = l
                }
            }
        }
    }
    //console.log(mar)
    return largest
};