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.
thinking:
(1)短邊決定水箱的有效高,底要盡可能的寬。
(2)典型的雙指針求解的題型。
(3)貪心的策略,哪條邊短,往裡收縮尋找下一條邊。
code:
class Solution {
public:
int maxArea(vector &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;
}
};
時間復雜度為O(n)
int area(vector::iterator &a,vector ::iterator &b) { return (b-a)*(*a>*b?*b:*a); } class Solution { public: int maxArea(vector &height) { int max_area=0; for(vector ::iterator i=height.begin()+1;i!=height.end();i++) { for(vector ::iterator j=height.begin();j!=i;j++) { max_area=max(max_area,area(j,i)); } } return max_area; } };