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; } };