leetcode——largestRectangleArea

题目描述

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 =10unit.

For example,
Given height =[2,1,5,6,2,3],

return10.

暴力：O(n^2)

#include <iostream>#include <vector>#include <string>#include <limits>#include <stack>using namespace std;class Solution {public:int largestRectangleArea(vector<int> &height) {int maxH = 0;int maxA = 0;for( int i= 0; i< height.size();i++){if( height[i] > maxH)maxH = height[i];}for( int h = 1; h<= maxH;h++){int maxLocal = 0;int maxT = 0;for( int i= 0; i< height.size();i++){if( height[i] >= h){maxT +=h;}else{if( maxLocal < maxT)maxLocal = maxT;maxT = 0;}}if( maxA < maxLocal){maxA = maxLocal;}}return maxA;}};int main(){string x = "()";vector<int > nn;nn.push_back(2);nn.push_back(1);nn.push_back(5);nn.push_back(6);nn.push_back(2);nn.push_back(3);//token.push_back("3");//token.push_back("*");Solution ss;cout<<ss.largestRectangleArea(nn);}

切头法：最坏O(n^2)；最好O(n);平均O(n^2);

基本思路：

每次添加一个队尾：1、如果此时队列中身高都低于队尾，则直接添加；

2、如果队列中有人比队尾高，那么高出来的这部分在队尾之后就没有计算价值了，因为队尾的身高限制了连通高度，此时可以收割；

3、收割：指将队伍中比队尾高的队员切头，使其身高等于队尾，在切头的过程中，计算其贡献值，更新max。像割麦子一样。

此法很容易改写为O(n),只需要将vector 换位 stack

class Solution {public:    int largestRectangleArea(vector<int> &height) {        //使用栈，但我偏不。。int max = 0;vector<int> cnt;//最终队尾为0，最终全部切为0；height.push_back(0);for( int i = 0; i< height.size() ; i++){int k = i-1;int local = 0;while( k >= 0 && cnt[k] >= height[i]){//计算将被切头部分的最大值if( (i - k) * cnt[ k ] > local)local =  (i - k) * cnt[ k ];//切头cnt[k] = height[i];k--;}if(max < local )max = local;//添加队尾cnt.push_back(height[i]);}        return max;    }};