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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.
我能想到的O(n^2)的算法。不过会出现超时。
C++实现如下：
#include<iostream> #include<vector> using namespace std; class Solution { public: int maxArea(vector<int> &height) { if(height.empty()||height.size()==1) return 0; int n=height.size(); int i,j; int maxArea=0; for(i=0; i<n-1; i++) { for(j=i+1; j<n; j++) { int tmp=min(height[i],height[j])*(j-i); if(maxArea<tmp) maxArea=tmp; } } return maxArea; } }; int main() { Solution s; vector<int> vec={2,4,1,3,0,6}; cout<<s.maxArea(vec)<<endl; }
参考：http://www.cnblogs.com/lichen782/p/leetcode_Container_With_Most_Water.html
O(n)的复杂度。保持两个指针i,j；分别指向长度数组的首尾。如果ai 小于aj，则移动i向后（i++）。反之，移动j向前（j--）。如果当前的area大于了所记录的area，替换之。这个想法的基础是，如果i的长度小于j，无论如何移动j，短板在i，不可能找到比当前记录的area更大的值了，只能通过移动i来找到新的可能的更大面积。
C++实现代码如下：
#include<iostream> #include<vector> using namespace std; class Solution { public: int maxArea(vector<int> &height) { if(height.empty()||height.size()==1) return 0; int n=height.size(); int i,j; int maxArea=0; i=0; j=n-1; while(i<j) { int tmp=min(height[i],height[j])*(j-i); if(maxArea<tmp) maxArea=tmp; if(height[i]<height[j]) i++; else j--; } return maxArea; } }; int main() { Solution s; vector<int> vec= {2,4,1,3,0,6}; cout<<s.maxArea(vec)<<endl; }
标签：style blog http io ar color os sp for
原文地址：http://www.cnblogs.com/wuchanming/p/4109111.html
