LeetCode 85. Maximal Rectangle&221. Maximal Square--动态规划

题目链接

221. Maximal Square

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

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

解：dp[i][j]表示以(i, j)为右下顶点的正方形的最长边长，那么我们所求的最大的面积则可以表示为max{ dp[i][j] }的平方，其中0<=i<height, 0<=j<width，易得转移方程：dp[i][j] = min{ dp[i-1][j], dp[i][j-1],
dp[i-1][j-1] }+1，可以在计算dp数组的同时维护一个maxLen（最长边长）变量，计算完成后得到的maxLen最终值平方就是答案，算法的复杂度为O(height*width)，代码：

class Solution {
public:
int maximalSquare(vector<vector<char> >& matrix) {
int height = matrix.size();
if (height == 0) return 0;
int width = matrix[0].size();
int** dp = new int*[height];
int maxLen = 0;
for (int i = 0; i < height; i++) dp[i] = new int[width];
for (int i = 0; i < height; i++) {
dp[i][0] = matrix[i][0]-'0';
maxLen = max(maxLen, dp[i][0]);
}
for (int i = 0; i < width; i++) {
dp[0][i] = matrix[0][i]-'0';
maxLen = max(maxLen, dp[0][i]);
}
for (int i = 1; i < height; i++) {
for (int j = 1; j < width; j++) {
if (matrix[i][j] == '0') dp[i][j] = 0;
else dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1])+1;
maxLen = max(maxLen, dp[i][j]);
}
}
return maxLen*maxLen;
}
};

题目链接

85. Maximal Rectangle

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

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 6.

解：和上题很相似，但是条件是矩形后情况就会比较复杂，对于(i, j)，我们需要分别考虑他的最大width（以(i,
j)为右顶点最长‘1‘线段长）和最大height（竖直的以(i, j)结尾的最长‘1’线段长），例如example中dpheight[2][3] = 2, dpwidth[2][3] = 4，而且最大矩形面积也不能简单的由最大长宽相乘获得，分别求得了dpheight和dpwidth后，可以以dpheight[i][j]向上遍历求得以(i, j)为右下顶点的每个矩形面积，也可以以dpwidth[i][j]向左遍历求得矩形面积，最终结果就是所有矩形面积的最大值了。

为了提高算法的效率，我们可以先只计算dpwidth，转移方程：dpwidth[i][j] = matrix[i][j] == '0' ? 0 : dpwidth[i][j-1]+1，然后对于每个(i, j)，动态地计算所有矩形面积，算法的时间复杂度O(width*height*height)，代码：

class Solution {
public:
int maximalRectangle(vector<vector<char> >& matrix) {
int height = matrix.size();
if (height == 0) return 0;
int width = matrix[0].size();
int** dpwidth = new int*[height];
for (int i = 0; i < height; i++) dpwidth[i] = new int[width];
int maxArea = 0;
for (int i = 0; i < height; i++) {
dpwidth[i][0] = matrix[i][0]-'0';
for (int t = i; t >= 0 && matrix[t][0] == '1'; t--) {
maxArea = max(maxArea, i-t+1);
}
}
for (int i = 0; i < height; i++) {
for (int j = 1; j < width; j++) {
dpwidth[i][j] = matrix[i][j] == '0' ? 0 : dpwidth[i][j-1]+1;
int wid = width;
for (int t = i; t >= 0 && matrix[t][j] == '1'; t--) {
wid = min(wid, dpwidth[t][j]);
maxAre
4000
a = max(maxArea, wid*(i-t+1));
}
}
}
return maxArea;
}
};