leetcode题解||Container With Most Water问题

problem：

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.

X轴为底，两个纵轴为变，求容器的容积，短边是瓶颈。

thinking:

(1)短边决定水箱的有效高，底要尽可能的宽。

（2）典型的双指针求解的题型。

（3）贪心的策略，哪条边短，往里收缩寻找下一条边。

code:

class Solution {
public:
int maxArea(vector<int> &height) {
// Start typing your C/C++ solution below
// DO 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;
}
};
时间复杂度为O（n）

暴力破解法：时间复杂度为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;

}
};