Container with most Water--leetcode

Topic:

GivenNnon-negative integersa1 ,a2 , ...,an , where each represents a point at coordinate (I,ai ).NVertical lines is drawn such that the both endpoints of lineIis 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.

idea: If the simplest way is to use brute force search, squared time complexity, but you can also use a simple method

Next we'll consider how to optimize. The idea is somewhat similar to the two Sum The second method of--clamp force. From the end of the array, each iteration to determine the number of left pointer and right pointer pointing to which large, if the left pointer small, meaning that the left to move right pointer is not possible to make the result better, because the bottleneck in the left pointer, moving right pointer will only be smaller, So this time we choose left pointer to move right. Conversely, the right pointer is selected to move left. In this process has been to maintain the largest volume. The code is as follows:

#include <iostream> #include <vector> #include <string> #include <stack>using namespace std;/*      Can fit up to water */int area (vector<int> &height, int i, int j) {int h = height[i]

Container with most Water--leetcode