leetcode — largest-rectangle-in-histogram

import java.util.Stack;

/**
*
* Source : https://oj.leetcode.com/problems/largest-rectangle-in-histogram/ *
*
* 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.
*
*                    6
*                  +---+
*               5  |   |
*              +---+   |
*              |   |   |
*              |   |   |
*              |   |   |     3
*              |   |   |   +---+
*        2     |   |   | 2 |   |
*      +---+   |   |   +---+   |
*      |   | 1 |   |   |   |   |
*      |   +---+   |   |   |   |
*      |   |   |   |   |   |   |
*      +---+---+---+---+---+---+
*
* Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
*
*
*                    6
*                  +---+
*               5  |   |
*              +-------|
*              |-------|
*              |-------|
*              |-------|     3
*              |-------|   +---+
*        2     |-------| 2 |   |
*      +---+   |-------|---+   |
*      |   | 1 |-------|   |   |
*      |   +---|-------|   |   |
*      |   |   |-------|   |   |
*      +---+---+---+---+---+---+
*
*
* The largest rectangle is shown in the shaded area, which has area = 10 unit.
*
* For example,
* Given height = [2,1,5,6,2,3],
* return 10.
*/
public class LargestRectangle {

/**
* 找到直方图中面积最大的矩形的面积
*
* 从左向右遍历数组
*      如果当前元素大于栈顶元素，说明当前是一个递增序列，将当前元素压入栈
*      如果当前元素小于栈顶元素，则pop出栈顶元素，计算当前栈顶元素和到当前位置的面积，和最大面积比较，更新最大面积，直到栈为空
*
*
* 边界条件
* 为了计算第一个元素，在栈底压入0
*
* @param arr
* @return
*/
public int findLargest (int[] arr) {
int result = 0;
Stack<Integer> stack = new Stack<Integer>();
stack.push(0);
for (int i = 0; i < arr.length; i++) {
if (stack.empty() || arr[i] >= arr[stack.peek()]) {
stack.push(i);
} else {
int cur = stack.pop();
result = Math.max(result, arr[cur] * (i - cur));
i --;
}
}
return result;
}

/**
* 相比于上面的方法，这里会将每一个递增序列前面所有的元素计算一遍，而不是仅仅计算当前递增序列
*
* @param arr
* @return
*/
public int findLargest1 (int[] arr) {
int res = 0;
for (int i = 0; i < arr.length; ++i) {
if (i + 1 < arr.length && arr[i] <= arr[i + 1]) {
continue;
}
int minH = arr[i];
for (int j = i; j >= 0; --j) {
minH = Math.min(minH, arr[j]);
int area = minH * (i - j + 1);
res = Math.max(res, area);
}
}
return res;
}

public static void main(String[] args) {
LargestRectangle largestRectangle = new LargestRectangle();
int[] arr = new int[]{2,1,5,6,2,3};
int[] arr1 = new int[]{2,1,5,6,5,2,3};
//        System.out.println(largestRectangle.findLargest(arr));
//        System.out.println(largestRectangle.findLargest(arr1));
System.out.println(largestRectangle.findLargest1(arr1));
}
}