HackerRank Largest Rectangle

题目链接


Problem Statement

There are N
buildings in a certain two-dimensional landscape. Each building has a height given byhi,i∈[1,N].
If you join K
adjacent buildings, they will form a solid rectangle of area
K×min(hi,hi+1,…,hi+k−1).

Given N
buildings, find the greatest such solid area formed by consecutive buildings.

Input Format

The first line contains N,
the number of buildings altogether.

The second line contains N
space-separated integers, each representing the height of a building.

Constraints
1≤N≤105
1≤hi≤106

Output Format

One integer representing the maximum area of rectangle formed.

Sample Input

5
1 2 3 4 5

Sample Output

9

Explanation

An illustration of the test case follows.

.

#include<stdio.h>			//Largest Rectangle
#include<string.h>
#include<stdlib.h>
#define MAXN 100001
#define MAXH 1000001
/*
算法思想：
对于数组中的每一个元素a[i]，计算以a[i]作为最小值的矩形的面积area，求出最大的面积即可。

为了计算以a[i]作为最小值的矩形的面积，我们需要知道a[i]之前第一个比a[i]小的元素索引（leftindex)以及a[i]之后第一个比a[i]小
的元素索引(rightindex),这样以[i]作为最小值的矩形的面积即为a[i]*(前后索引之间的距离)

利用数据结构栈（栈内元素递增），将数组元素依次与栈顶元素比较，当栈空时，直接入栈，当栈顶元素小于等于数组元素时，数组元素
直接入栈，否则，栈顶元素出栈直至栈顶元素小于等于数组元素，再将数组元素入栈。当所有元素都遍历过之后，若栈非空，则一次弹栈，
并计算相应的面积，直至栈空即可

为了能够知道a[i]的leftindex和rightindex，在数组元素入栈时需要记录元素的索引
*/

int a[MAXN], stack[MAXN];
int top = -1;
int area[MAXN];			//记录以元素a[i]为最小值的矩形的面积
int index[MAXN];		//记录栈中元素的原来序号(即A[i]的i)

int FindMax(int n)
{
int i, leftindex, rightindex, max=0;
for (i = 0; i < n; i++)
{
if (top < 0 || (top >= 0 && stack[top]<=a[i]))		//直接入栈
{
stack[++top] = a[i];
index[top] = i;
}
else
{
while (top >= 0 && stack[top] > a[i])
{
//计算以栈顶元素作为最小值的矩形的面积
//top>0时，right即为i，leftindex为栈顶元素之前元素的索引，即index[top-1]
//top==0时，rightindex和rightindex之间的距离即为i
area[index[top]] = stack[top] * (top > 0 ? (i - index[top - 1] - 1) : i);
if (max < area[index[top]]) max = area[index[top]];		//找最大
top--;
}
stack[++top] = a[i];
index[top] = i;
}
}
while (top >= 0)		//数组元素遍历完毕，若栈非空，则将栈中元素依次出栈，并计算相应的面积
{
area[index[top]] = stack[top] * (top > 0 ? (i - index[top - 1] - 1) : i);
if (max < area[index[top]]) max = area[index[top]];
top--;
}
return max;
}

int main()
{
int n, i;
scanf("%d", &n);
for (i = 0; i < n; i++)
scanf("%d", a + i);
printf("%d\n", FindMax(n));
return 0;
}