LeetCode算法题——11. Container With Most Water

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

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1  算法思想，计算每个面积求出最大值

class Solution {

public:

    int maxArea(vector<int>& height) {

        int areaMax=-1;

        for(int i=0;i<height.size()-1;i++){

            for(int j=i+1;j<height.size();j++){

                areaMax=max(areaMax,min(height[i],height[j])*(j-i));               

            }

        }

        return areaMax;

    }

};

在大数据情况下超时

2 改进算法思想

现假定取得最大面积的坐标为（i,height[i]）（j,height[j]），其中i<j,

则areaMax=max(areaMax,min(height[i],height[j])*(j-i)),由此可推知在j的右端没有一条线比它高，设k>j,且height[k]>height[j],min(height[i],height[k])= min(height[i],height[j]),则areaMax'= max(areaMax,min(height[i],height[j])*(k-i))>areaMax,与原假设矛盾。

同理可知，i的左端没有一条线比它低

#include <iostream>

#include <vector>

#include "math.h"

using namespace std;

class Solution {

public:

    int maxArea(vector<int>& height) {

        int areaMax=-1;

        int i=0,j=height.size()-1;

//        for(int i=0;i<height.size()-1;i++){

//            for(int j=i+1;j<height.size();j++){

//                areaMax=max(areaMax,min(height[i],height[j])*(j-i));               

//            }

//        }

        while(i<j){

            areaMax=max(areaMax,min(height[i],height[j])*(j-i));

            if(height[i]>height[j]){

                j--;

            }else{

                i++;

            }

        }

        return areaMax;

    }

};

int main(){

    vector<int> v;

    v.push_back(1);

    v.push_back(2);

    v.push_back(1);

    Solution sol;

    int result=sol.maxArea(v);

    cout<<result;

    return 0;

}