Description Title: Given a matrix N✖N matrix, this matrix, only two kinds of values of 0 and 1, all the border returns the maximum length of a square side length of 1.
E.g:
{0,1,1,1,1},
{0,1,0,0,1},
{0,1,0,0,1},
{0,1,1,1,1},
{0,1,0,1,1}
其中，边框全是1的最大正方形的大小是4*4，故返回4。
This question before I wrote a search method using violence to solve, but a relatively large time complexity O (N ^ 4) may time out, did not read the small partner may have a look: violence Search version
Here we first of this matrix pretreatment, somewhat similar dynamic programming idea is to create aN✖N✖2The three-dimensional array, wherein a predetermined, if each element is 1, then look at the right and lower sides, respectively, the number of elementscontinuous1, and then use the position number 0 of the final three-dimensional memory array to the right of the (Including itself) Is the number 1, No. 1 of the last dimension able to save the side position (Including itself) Number 1. There should be little skill is initialized from the last element of the last row (i.e., [N-1] [N-1]) forward. Then we check the auxiliary three-dimensional array meets a particular number (eg n-order) to order - check method is very simple, just look at the current element (this is the top-left corner of the border) and the right side whether lower side at least n 1, if the right lower vertex side at least n 1, and if the right side of the lower left apex of at least n-1.
Look at the code it:
#include<iostream>
#define N 5
using namespace std;
int solve(int matrix[][N]){
int helper[N][N][2]={0}; //申请一个辅助数组空间
int r = N-1,c = N-1; //r,c为要预处理的当前位置，从最后一个元素开始
//先对最后一行初始化
for(;c>=0;c--){
if(matrix[r][c] == 1){ //前提得是1
if(c == N-1){ //如果是最后一列
helper[r][c][0] = 1;
helper[r][c][1] = 1;
}
else{
helper[r][c][0] = helper[r][c+1][0] + 1;
helper[r][c][1] = 1;
}
}
}
//再对前面的预处理
//防止越界
if(N-2 >= 0){
r = N-2;
for(;r>=0;r--){
for(c = N-1;c>=0;c--){
if(matrix[r][c] == 1){ //前提得是1
if(c == N-1){ //如果是最后一列
helper[r][c][0] = 1;
helper[r][c][1] = helper[r+1][c][1] + 1;
}
else{
helper[r][c][0] = helper[r][c+1][0] + 1;
helper[r][c][1] = helper[r+1][c][1] + 1;
}
}
}
}
}
int n = N; //初始阶数为N
while(n>=0){
for(int i=0;i<N;i++){
if(i + n > N) break;
for(int j=0;j<N;j++){
if(j + n > N) break;
//check一下左上顶点的右边和下边，右上顶点的下边以及左下顶点的右边是否满足n阶
if(helper[i][j][0] >= n && helper[i][j][1] >= n && helper[i][j+n-1][1] >= n && helper[i+n-1][j][0] >= n){
return n;
}
}
}
n--;
}
return n; //全为0的情况
}
int main(){
int matrix[N][N] = {
{0,1,1,1,1},
{0,1,0,0,1},
{0,1,0,0,1},
{0,1,1,1,1},
{0,1,0,1,1}
};
int res = solve(matrix);
cout<<"result: "<<res;
}
Run the result is the same:
we also welcome to test the absence of flaws, I measured the time in addition to matrix is empty except for a small problem, others are normal, if there is a problem please generously pointed out, oh.
Time Complexity Analysis: initialization auxiliary array is N² (N + N²), and when the inspection is N³ (while: N, for✖2: N²), it is only necessary complexity of O (N³) a.
