Maximal Rectangle
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing all ones and return its area.class Solution {
public:
int maximalRectangle(vector<vector<char> > &matrix) {
int m = matrix.size();
if(m == 0){
return 0;
}
int n = matrix[0].size();
if(n==0){
return 0;
}
int** d=new int*[m + 1];
for(int i=0; i<=m; i++){
d[i]=new int[n + 1];
}
for(int i=0; i<=m; i++){
d[i][0]=0;
}
for(int i=0; i<=n; i++){
d[0][i]=0;
}
for(int i=1; i<=m; i++){
for(int j=1; j<=n; j++){
d[i][j]=max(d[i-1][j], d[i][j-1]);
if(matrix[i-1][j-1]=='1'){
//计算以i-1,j-1元素为右下角的最大全1矩阵的面积
//找到i方向最小的i
int minI = i-1;
while(minI >=0 && matrix[minI][j-1]=='1') minI--;
minI = max(minI, 0);
//找到j方向最小的j
int minJ = j-1;
while(minJ >=0 && matrix[i-1][minJ]=='1') minJ--;
minJ = max(minJ, 0);
if((i - minI)*(j - minJ) > d[i][j]){
int maxArea = 0;
for(int tempI = minI; tempI < i; tempI++){
for(int tempJ = minJ; tempJ < j; tempJ++){
int area = getArea(matrix, tempI, i-1, tempJ, j-1);
if(area!=0){
maxArea = max(area, maxArea);
break;
}
}
}
d[i][j]=max(d[i][j], maxArea);
}
}
}
}
int result = d[m][n];
for(int i=0; i<=m; i++){
delete[] d[i];
}
delete[] d;
return result;
}
private:
int max(int a, int b){
return a>b?a:b;
}
int getArea(vector<vector<char>>& matrix, int startI, int endI, int startJ, int endJ){
bool allOne = true;
for(int i=startI; i<=endI; i++){
for(int j=startJ; j<=endJ; j++){
if(matrix[i][j]=='0'){
allOne = false;
break;
}
}
if(!allOne){
break;
}
}
if(allOne){
return (endI - startI + 1) * (endJ - startJ + 1);
}else{
return 0;
}
}
};居然也能通过，但运行时间是874ms，显然是不对的，因为其最坏情况运行时间复杂度为O(m2*n2)class Solution {
public:
int maximalRectangle(vector<vector<char> > &matrix) {
int m = matrix.size();
if(m == 0){
return 0;
}
int n = matrix[0].size();
if(n==0){
return 0;
}
//动态规划数组
int** d=new int*[m + 1];
for(int i=0; i<=m; i++){
d[i]=new int[n + 1];
}
for(int i=0; i<=m; i++){
d[i][0]=0;
}
for(int i=0; i<=n; i++){
d[0][i]=0;
}
//记录全1的高度
int* h = new int[n + 1];
for(int i=0; i<=n; i++){
h[i] = 0;
}
for(int i=1; i<=m; i++){
for(int j=1; j<=n; j++){
d[i][j]=max(d[i-1][j], d[i][j-1]);
if(matrix[i-1][j-1]=='1'){
h[j]++;
int maxArea = h[j];
int minH = h[j];
for(int k=j-1; k>0; k--){
if(h[k] == 0){
break;
}
minH = minH > h[k] ? h[k] : minH;
maxArea = max(maxArea, minH * (j - k + 1));
}
d[i][j] = max(d[i][j], maxArea);
}else{
h[j]=0;
}
}
}
int result = d[m][n];
for(int i=0; i<=m; i++){
delete[] d[i];
}
delete[] d;
delete[] h;
return result;
}
private:
int max(int a, int b){
return a>b?a:b;
}
};这样，算法的时间复杂度降到了O(m*n*n)，这是坏的情况。在LeetCode中运行时间降到了28ms。class Solution {
public:
int maximalRectangle(vector<vector<char> > &matrix) {
int m = matrix.size();
if(m == 0){
return 0;
}
int n = matrix[0].size();
if(n==0){
return 0;
}
//记录全1的高度
int* h = new int[n];
for(int i=0; i<n; i++){
h[i] = 0;
}
int result = 0;
for(int i=0; i<m; i++){
for(int j=0; j<n; j++){
if(matrix[i][j]=='1'){
h[j]++;
int maxArea = h[j];
int minH = h[j];
for(int k=j-1; k>=0; k--){
if(h[k] == 0){
break;
}
minH = minH > h[k] ? h[k] : minH;
maxArea = max(maxArea, minH * (j - k + 1));
}
result = max(result, maxArea);
}else{
h[j]=0;
}
}
}
return result;
}
private:
int max(int a, int b){
return a>b?a:b;
}
};这样，时间复杂度虽然没有变，但是省去了很大的空间开销，从而也节省了时间。LeetCode的运行时间为19msclass Solution {
public:
int maximalRectangle(vector<vector<char> > &matrix) {
int m = matrix.size();
if (m == 0){
return 0;
}
int n = matrix[0].size();
if (n == 0){
return 0;
}
int result = 0;
//记录全1的高度
int* h = new int[n+1]; //多申请一个空间，稍后会知道他的好处
for (int i = 0; i<=n; i++){
h[i] = 0;
}
for (int i = 0; i<m; i++){
stack<int> s;
int tempMax = 0; //某一行最大的area
bool hCounted = false;
for (int j = 0; j<=n; j++){
if(!hCounted&&j!=n){
if (matrix[i][j] == '1'){
h[j]++;
}
else{
h[j] = 0;
}
hCounted = true;
}
if (s.empty() || h[s.top()]<h[j]){ //入栈
s.push(j);
hCounted = false;
}
else{ //h多申请了一个空间，并赋值为0，保证会最终会执行到此步
int temp = s.top();
s.pop();
tempMax = max(tempMax, h[temp] * (s.empty() ? j : (j - s.top() - 1)));
j--; //这里j不变，表示找出所有大于当前的，并出栈
}
}
result = max(result, tempMax);
}
delete[] h;
return result;
}
private:
int max(int a, int b){
return a>b ? a : b;
}
};原文地址：http://blog.csdn.net/kangrydotnet/article/details/44786169
