Leetcode | | Container with most water issues

Problem:

Given n non-negative integers a1, A2, ..., an, where each represents a point at coordinate (I, AI). n vertical lines is drawn such, the and the other endpoints of line I am at (i, AI) and (i, 0). Find lines, which together with X-axis forms a container, such that the container contains the most water. Note:you may not slant the container.

The x-axis is the bottom, two longitudinal axes are changed, the volume of the container is obtained, and the short edge is the bottleneck.

Thinking:

(1) The short side determines the effective height of the tank, the bottom to be as wide as possible.

(2) A typical double-pointer solution.

(3) Greedy strategy, which side is short, narrowing in the search for the next edge.

Code

Class Solution {public:    int Maxarea (vector<int> &height) {        //Start Typing your C + + solution Below
   //does not write int main () function        int i = 0;        Int J = height.size ()-1;                int ret = 0;        while (I < j)        {            int area = (j-i) * min (Height[i], height[j]);            RET = max (ret, area);                        if (Height[i] <= height[j])                i++;            else                j--;        }                return ret;    }};

Time Complexity of O (n)

Brute force method: Time Complexity O (n*n)

int area (Vector<int>::iterator &a,vector<int>::iterator &b) {    return (b-a) * (*a>*b?*b:*a) ;} Class Solution {public:    int Maxarea (vector<int> &height) {        int max_area=0;        For (Vector<int>::iterator I=height.begin () +1;i!=height.end (), i++)        {for            (VECTOR<INT>:: Iterator J=height.begin (); j!=i;j++)            {                Max_area=max (Max_area,area (j,i));            }        }        return max_area;}    ;

Leetcode | | Container with most water issues