LeetCode No.84 Largest Rectangle in Histogram
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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.
Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3]
.
The largest rectangle is shown in the shaded area, which has area = 10
unit.
For example,
Given heights = [2,1,5,6,2,3]
,
return 10
.
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这道题目的大意是:求出给定柱形图的最大长方形面积。
刚看到这道题的时候,我想应该不是很难,只要确定一个柱体为中点,然后记录不小于该柱体高度的往左走的最小left和往右走的最大right,然后求(right-left+1)*height的最大值就行了,这种做法的时间复杂度为O(N^2)。但是题目难度级别是hard,如果O(N^2)的算法都能过的话那不就很尴尬了,果不其然,TLE了。那有没有O(N*logN)甚至是O(N)的算法可以解决这个问题呢?
答案是有的,先放一个最小的0在高度序列的后面(预处理最后操作)从左往右扫一遍柱体,用一个vector记录这个严格递增的高度序列对应的下标,但出现不是严格递增的高度时,记录下此时最大矩阵,公式为:ans = max ( ans , heights[j] * ( hs.empty() ? i : i - hs.back() - 1 ) ) ,这样的时间复杂度便是O(N)了。
class Solution {public: int largestRectangleArea(vector<int>& heights) { heights.push_back ( 0 ) ; vector <int> hs ; int i = 0 , ans = 0 ; while ( i < heights.size() ) { if ( hs.empty() || heights[i] > heights[hs.back()] ) { hs.push_back ( i ++ ) ; } else { int j = hs.back () ; hs.pop_back () ; ans = max ( ans , heights[j] * ( hs.empty() ? i : i - hs.back() - 1 ) ) ; } } return ans ; }};
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