给定非负整数数组 heights ,数组中的数字用来表示柱状图中各个柱子的高度。每个柱子彼此相邻,且宽度为 1 。
求在该柱状图中,能够勾勒出来的矩形的最大面积。
示例 1:输入:heights = [2,1,5,6,2,3] 输出:10
解释:最大的矩形为图中红色区域,面积为 10
示例 2:输入: heights = [2,4] 输出: 4
提示:1 <= heights.length <=105
0 <= heights[i] <= 104
注意:本题与主站 84 题相同:
1、单调栈;时间复杂度O(n),空间复杂度O(n)
func largestRectangleArea(heights []int) int {
n := len(heights)
res := 0
left := make([]int, n)
right := make([]int, n)
stack := make([]int, 0)
for i := 0; i < n; i++ {
for len(stack) > 0 && heights[stack[len(stack)-1]] >= heights[i] {
stack = stack[:len(stack)-1]
}
if len(stack) == 0 {
left[i] = -1
} else {
left[i] = stack[len(stack)-1]
}
stack = append(stack, i)
}
stack = make([]int, 0)
for i := n - 1; i >= 0; i-- {
for len(stack) > 0 && heights[stack[len(stack)-1]] >= heights[i] {
stack = stack[:len(stack)-1]
}
if len(stack) == 0 {
right[i] = n
} else {
right[i] = stack[len(stack)-1]
}
stack = append(stack, i)
}
for i := 0; i < n; i++ {
res = max(res, heights[i]*(right[i]-left[i]-1))
}
return res
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
2、单调栈;时间复杂度O(n),空间复杂度O(n)
func largestRectangleArea(heights []int) int {
n := len(heights)
res := 0
left := make([]int, n)
right := make([]int, n)
stack := make([]int, 0)
for i := 0; i < n; i++ {
right[i] = n
}
for i := 0; i < n; i++ {
for len(stack) > 0 && heights[stack[len(stack)-1]] >= heights[i] {
right[stack[len(stack)-1]] = i
stack = stack[:len(stack)-1]
}
if len(stack) == 0 {
left[i] = -1
} else {
left[i] = stack[len(stack)-1]
}
stack = append(stack, i)
}
for i := 0; i < n; i++ {
res = max(res, heights[i]*(right[i]-left[i]-1))
}
return res
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
3、单调栈;时间复杂度O(n),空间复杂度O(n)
func largestRectangleArea(heights []int) int {
heights = append([]int{0}, heights...)
heights = append(heights, 0)
n := len(heights)
res := 0
stack := make([]int, 0)
for i := 0; i < n; i++ {
// 递增栈
for len(stack) > 0 && heights[stack[len(stack)-1]] > heights[i] {
height := heights[stack[len(stack)-1]]
stack = stack[:len(stack)-1]
width := i - stack[len(stack)-1] - 1
res = max(res, height*width)
}
stack = append(stack, i)
}
return res
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
Hard题目,题目类似有
leetcode 84.柱状图中最大的矩形、
leetcode 1793.好子数组的最大分数、
leetcode 1856.子数组最小乘积的最大值
页面更新:2024-03-30
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