"Dynamic planning" leetcode-maximal Square

Title:

Maximal Square 

Given a 2D binary matrix filled with 0 's and 1 ' s, find the largest square containing all 1 's and return its area.

For example, given the following matrix:

1 1 1 1 11 0 0 1 0

Return 4.

Analysis:

Solve with dynamic programming. Create a class node. The member variable left in node has a number of 1 (containing the point itself), a few 1 (containing the point itself), and a side length (the point in the lower right corner of the square) of the corresponding maximum square of the maxsize. If a point is ' 0 ', then its corresponding node is (0,0,0).

1. Use variable res to record the edge length of the largest square.

2, first of all, the input matrix is processed in the upper left corner, the first row and the first column, to find out the node value of these locations.

3. Go through the rest of the matrix in turn and find the node value for each point. and update Res.

4, return to Res*res.

Class node{Public:int left,up,maxsize; Node (): Left (0), up (0), maxsize (0) {} node (Int. a,int b,int C): Left (a), up (b), maxsize (c) {}};class Solution {Public:int maxi  Malsquare (vector<vector<char>>& matrix) {if (Matrix.empty () | | matrix[0].empty ()) return                    0;        int rows=matrix.size (), cols=matrix[0].size ();        int res=0;                Vector<vector<node>> DP (rows,vector<node> (COLS));            if (matrix[0][0]== ' 1 ') {Res=1;        Dp[0][0]=node (1,1,1);                } for (int j=1;j<cols;++j) {if (matrix[0][j]== ' 1 ') {Res=1;            Dp[0][j]=node (dp[0][j-1].left+1,1,1);                }} for (int i=1;i<rows;++i) {if (matrix[i][0]== ' 1 ') {Res=1;            Dp[i][0]=node (1,dp[i-1][0].up+1,1); }} for (int i=1;i<rows;++i) {T j=1;j<cols;++j) {if (matrix[i][j]== ' 1 ') {DP[I][J].L                     eft=dp[i][j-1].left+1;                     dp[i][j].up=dp[i-1][j].up+1;                     if (matrix[i-1][j-1]!= ' 1 ') dp[i][j].maxsize=1;                         else {int tmp=min (dp[i-1][j-1].maxsize+1,dp[i][j].left);                          Tmp=min (tmp,dp[i][j].up);                     dp[i][j].maxsize=tmp;                  } res=max (Res,dp[i][j].maxsize);    }}} return res*res; }};

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"Dynamic planning" leetcode-maximal Square