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.
思考：想用分治思想来解决这个问题，（a1,...,an-1) 和（a1,...,an-1,an）答案之间的关系是
S（a1,...,an-1) = max {S(a1,...,an-1), max(an作为右边，右边分别为a1到an-1）}
实现算法如下。
1 int maxArea(int* height, int heightSize) { 2 if(heightSize == 2) 3 return ((height[heightSize-1]>=height[heightSize-2])?height[heightSize-2]:height[heightSize-1]); 4 5 int tmp,compare1,compare2,result,i; 6 7 compare1 = (heightSize - (heightSize-1))*((height[heightSize-1]>=height[heightSize-2])?height[heightSize-2]:height[heightSize-1]); 8 for(i=heightSize-2; i>0; i--) { 9 tmp = (heightSize - i)*((height[heightSize-1]>=height[i-1])?height[i-1]:height[heightSize-1]); 10 if(tmp>compare1) compare1 = tmp; 11 } 12 13 compare2 = maxArea(height, heightSize-1); 14 if(compare2>compare1) 15 result = compare2; 16 else 17 result = compare1; 18 19 return result; 20 }
可是运行结果是时间超时了，分析下运行时间，递归深度为n,至顶而下，每一层的操作数分别为，n，n-1,n-2,...,2.为O(n平方）。超出时间。网上看了下这个博客的思路，自己实现了，如下。
http://bangbingsyb.blogspot.tw/2014/11/leetcode-container-with-most-water.html
1 int maxArea(int* height, int heightSize) { 2 3 int index1 = 1; 4 int index2 = heightSize; 5 int max = ((height[index1-1]>=height[index2-1])?height[index2-1]:height[index1-1])*(index2-index1); 6 int i,j,tmp; 7 8 i = 1; 9 j = heightSize; 10 while(j-i) { 11 if(height[j-1]>=height[i-1]) 12 i++; 13 else 14 j--; 15 tmp = ((height[i-1]>=height[j-1])?height[j-1]:height[i-1])*(j-i); 16 if(tmp>max) max = tmp; 17 } 18 19 return max; 20 }