leetcode 85.最大矩形
题干
给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。
示例 1:
输入:matrix = [[“1”,“0”,“1”,“0”,“0”],[“1”,“0”,“1”,“1”,“1”],[“1”,“1”,“1”,“1”,“1”],[“1”,“0”,“0”,“1”,“0”]]
输出:6
解释:最大矩形如上图所示。
示例 2:
输入:matrix = []
输出:0
示例 3:
输入:matrix = [[“0”]]
输出:0
示例 4:
输入:matrix = [[“1”]]
输出:1
示例 5:
输入:matrix = [[“0”,“0”]]
输出:0
提示:
rows == matrix.length
cols == matrix[0].length
0 <= row, cols <= 200
matrix[i][j] 为 ‘0’ 或 ‘1’
题解
遍历矩阵,记录每个1右侧连续1的个数(包括自己),然后将1的坐标存入队列,
然后遍历队列中每个1的坐标,检查以这个1为左上角时所能形成的面积最大的矩形,
策略为检查下方行的1的个数,不断取 已遍历的行数 * 已出现单行1出现的最少次数 的最大值为面积
class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
int rows = matrix.size();
if(rows == 0){
return 0;
}
int cols = matrix[0].size();
int maxRectangleArea = 0;
vector<vector<int> > rightContinuousOneCount(rows,vector<int>(cols,0));
queue<pair<int,int> > onePosition;
for(int i = 0 ; i < rows ; ++i){
for(int j = 0 ; j < cols ; ++j){
if(matrix[i][j] == '1'){
onePosition.push({
i,j});
int tempCols = j + 1;
int tempRightContinuousOneCount = 1;
while(tempCols < cols && matrix[i][tempCols] == '1'){
tempRightContinuousOneCount++;
tempCols++;
}
rightContinuousOneCount[i][j] = tempRightContinuousOneCount;
}
}
}
while(!onePosition.empty()){
int currentRow = onePosition.front().first;
int currentCol = onePosition.front().second;
onePosition.pop();
int tempMaxArea = rightContinuousOneCount[currentRow][currentCol];
int minRowLength = tempMaxArea;
int rowsCount = 1;
currentRow++;
rowsCount++;
while(currentRow < rows && rightContinuousOneCount[currentRow][currentCol] > 0){
minRowLength = min(minRowLength,rightContinuousOneCount[currentRow][currentCol]);
tempMaxArea = max(tempMaxArea,rowsCount * minRowLength);
currentRow++;
rowsCount++;
}
maxRectangleArea = max(maxRectangleArea,tempMaxArea);
}
return maxRectangleArea;
}
};