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
- 1 <= row, cols <= 200
- matrix[i][j] 为 ‘0’ 或 ‘1’
解题：
困难的不会做，想了一会儿，还是直接看答案官方解答：
1.使用柱状图的优化暴力方法
func maximalRectangle(matrix [][]byte) (ans int) {
if len(matrix) == 0 {
return
}
m, n := len(matrix), len(matrix[0])
left := make([][]int, m)
for i, row := range matrix {
left[i] = make([]int, n)
for j, v := range row {
if v == '0' {
continue
}
if j == 0 {
left[i][j] = 1
} else {
left[i][j] = left[i][j-1] + 1
}
}
}
for i, row := range matrix {
for j, v := range row {
if v == '0' {
continue
}
width := left[i][j]
area := width
for k := i - 1; k >= 0; k-- {
width = min(width, left[k][j])
area = max(area, (i-k+1)*width)
}
ans = max(ans, area)
}
}
return
}
func min(a, b int) int {
if a < b {
return a
}
return b
}
func max(a, b int) int {
if a > b {
return a
}
return b
}2.单调栈
func maximalRectangle(matrix [][]byte) (ans int) {
if len(matrix) == 0 {
return
}
m, n := len(matrix), len(matrix[0])
left := make([][]int, m)
for i, row := range matrix {
left[i] = make([]int, n)
for j, v := range row {
if v == '0' {
continue
}
if j == 0 {
left[i][j] = 1
} else {
left[i][j] = left[i][j-1] + 1
}
}
}
for j := 0; j < n; j++ { // 对于每一列，使用基于柱状图的方法
up := make([]int, m)
down := make([]int, m)
stk := []int{}
for i, l := range left {
for len(stk) > 0 && left[stk[len(stk)-1]][j] >= l[j] {
stk = stk[:len(stk)-1]
}
up[i] = -1
if len(stk) > 0 {
up[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
stk = nil
for i := m - 1; i >= 0; i-- {
for len(stk) > 0 && left[stk[len(stk)-1]][j] >= left[i][j] {
stk = stk[:len(stk)-1]
}
down[i] = m
if len(stk) > 0 {
down[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
for i, l := range left {
height := down[i] - up[i] - 1
area := height * l[j]
ans = max(ans, area)
}
}
return
}
func max(a, b int) int {
if a > b {
return a
}
return b
}