添加内容

website/content/ChapterFour/0084.Largest-Rectangle-in-Histogram.md

@@ -0,0 +1,75 @@
+# [84. Largest Rectangle in Histogram](https://leetcode.com/problems/largest-rectangle-in-histogram/)
+
+## 题目
+
+Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
+
+ ![](https://assets.leetcode.com/uploads/2018/10/12/histogram.png)
+
+
+Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
+
+![](https://assets.leetcode.com/uploads/2018/10/12/histogram_area.png)
+
+
+The largest rectangle is shown in the shaded area, which has area = 10 unit.
+
+ 
+
+**Example**:
+
+```
+
+Input: [2,1,5,6,2,3]
+Output: 10
+
+```
+
+
+## 题目大意
+
+给出每个直方图的高度，要求在这些直方图之中找到面积最大的矩形，输出矩形的面积。
+
+
+## 解题思路
+
+用单调栈依次保存直方图的高度下标，一旦出现高度比栈顶元素小的情况就取出栈顶元素，单独计算一下这个栈顶元素的矩形的高度。然后停在这里(外层循环中的 i--，再 ++，就相当于停在这里了)，继续取出当前最大栈顶的前一个元素，即连续弹出 2 个最大的，以稍小的一个作为矩形的边，宽就是 2 计算面积…………如果停在这里的下标代表的高度一直比栈里面的元素小，就一直弹出，取出最后一个比当前下标大的高度作为矩形的边。宽就是最后一个比当前下标大的高度和当前下标 i 的差值。计算出面积以后不断的更新 maxArea 即可。
+
+## 代码
+
+```go
+
+package leetcode
+
+import "fmt"
+
+func largestRectangleArea(heights []int) int {
+	maxArea, stack, height := 0, []int{}, 0
+	for i := 0; i <= len(heights); i++ {
+		if i == len(heights) {
+			height = 0
+		} else {
+			height = heights[i]
+		}
+		if len(stack) == 0 || height >= heights[stack[len(stack)-1]] {
+			stack = append(stack, i)
+		} else {
+			tmp := stack[len(stack)-1]
+			fmt.Printf("1. tmp = %v stack = %v\n", tmp, stack)
+			stack = stack[:len(stack)-1]
+			length := 0
+			if len(stack) == 0 {
+				length = i
+			} else {
+				length = i - 1 - stack[len(stack)-1]
+				fmt.Printf("2. length = %v stack = %v i = %v\n", length, stack, i)
+			}
+			maxArea = max(maxArea, heights[tmp]*length)
+			fmt.Printf("3. maxArea = %v heights[tmp]*length = %v\n", maxArea, heights[tmp]*length)
+			i--
+		}
+	}
+	return maxArea
+}
+
+```