85. Maximal Rectangle

题目：

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 01 0 1 1 11 1 1 1 11 0 0 1 0

Return 6.

解法1 动态规划：

最大面积= max（(right(i, j) - left(i, j)) * height(i, j) ）

其中height(i, j) 是以m(i, j)为起点的高。

right(i, j)是以height(i, j)为高的右边界，同理left(i, j)。

TIPS:确定合适的dp[i][j]非常重要。

class Solution {public:    int maximalRectangle(vector<vector<char>>& matrix) {        if (<span style="color:#ff0000;">matrix.empty()</span>) return 0;//注意输入是否为空        const int m = matrix.size() + 1;    const int n = matrix[0].size() ;    vector<vector<int>> left(m, vector<int>(n, 0));    vector<vector<int>> right(m, vector<int>(n, n));    vector<vector<int>> height(m, vector<int>(n, 0));    int ans = 0;    for(int i = 1; i < m; i++)    {        int cur_left = 0, cur_right = n;        for(int j = n - 1; j >= 0; j--)//注意j的初始值        {            if(matrix[i - 1][j] == '0')            {                height[i][j] = 0;                cur_right =  j;            }            else            {                height[i][j] = height[i - 1][j] + 1;                right[i][j] = min(right[i - 1][j], cur_right);            }        }        for(int j = 0; j < n; j++)        {            if(matrix[i - 1][j] == '0')                cur_left = j + 1;//注意要加一            else                left[i][j] = max(left[i - 1][j], cur_left);            ans = max(ans, (right[i][j] - left[i][j]) * height[i][j]);        }    }    return ans;    }};

空间复杂度o(m * n), 时间复杂度 o(m * n) ； 我们知道dp问题可以降维，下面是空间复杂度o(n)的代码

int maximalRectangle(vector<vector<char>>& matrix) {if (matrix.empty()) return 0;const int m = matrix.size();const int n = matrix[0].size();/*int left[n] = {0}, right[n], height[n];fill_n(left, n, 0); fill_n(right, n, n); fill_n(height, n, 0);*/vector<int> left(n, 0), right(n, n), height(n, 0);int maxA = 0;for (int i = 0; i<m; i++) {int cur_left = 0, cur_right = n;for (int j = 0; j<n; j++) { // compute height (can do this from either side)if (matrix[i][j] == '1') height[j]++;else height[j] = 0;}for (int j = 0; j<n; j++) { // compute left (from left to right)if (matrix[i][j] == '1') left[j] = max(left[j], cur_left);else { left[j] = 0; cur_left = j + 1; }}// compute right (from right to left)for (int j = n - 1; j >= 0; j--) {if (matrix[i][j] == '1') right[j] = min(right[j], cur_right);else { right[j] = n; cur_right = j; }}// compute the area of rectangle (can do this from either side)for (int j = 0; j<n; j++)maxA = max(maxA, (right[j] - left[j])*height[j]);for (int j = 0; j < n; j++){cout << left[j] << " ";if (j == n - 1)cout << endl;}for (int j = 0; j < n; j++){cout << right[j] << " ";if (j == n - 1)cout << endl;}cout << endl;}return maxA;}

解法2：栈+动态规划

待续