【LeetCode】011 Container With Most Water

题目：

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.

题解：

　　暴力解即可，注意这个面积是长方形的，我刚开始犯傻给写成梯形了。。。两个思路：两个for循环遍历，结果就是TLE了。另一个就是头尾指针遍历法，时间复杂度降为O(n).以后遇到这种需要遍历的，要先想到头尾指针法，而不是两个for循环(其实也是双指针，只是位置不一样)。同时，与之前遇到的数组问题一样，小幅度的优化就是在大循环内就跳过重复项

Solution 1 (TLE)

class Solution {
public:
    int maxArea(vector<int>& height) {
        int area = , n = height.size();
        for(int i=; i<n-; i++) {
            for(int j=i+; j<n; j++) {
                int new_area = (j-i)*min(height[i], height[j]);
                if(height[i]== || height[j]==) new_area = ;
                area = max(area, new_area);
            }
        }
        return area;
    }
};

　　

Solution 2

class Solution {
public:
    int maxArea(vector<int>& height) {
        int area = , i = , j = height.size() - ;
        while (i < j) {
            area = max(area, min(height[i], height[j]) * (j - i));
            height[i] < height[j] ? ++i : --j;
        }
        return area;
    }
};

Solution 3

class Solution {
public:
    int maxArea(vector<int>& height) {
        int area = ;
        int i = , j = height.size() - ;
        while (i < j) {
             int h = min(height[i], height[j]);
            area = max(area, (j - i) * h);
            while (height[i] <= h && i < j) i++;
            while (height[j] <= h && i < j) j--;
        }
        return area;
    }
};