第三周：11. Container With Most Water

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.

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我解这道题的思路是：将一条线段与其他线段都做遍历，找出最大值，其中代码如下：

int maxArea(int* height, int heightSize) {    int i,j;int result=0;int temp;for(i=0;i<heightSize;i++)for(j=i+1;j<heightSize;j++){if(height[i]<height[j])temp=height[i]*(j-i);else temp=height[j]*(j-i);if(result<temp)result=temp;}return result;}

显然这个方法会出现超时：

于是在网上看别人的解题思路：先拿最左边的线段和最右边的线段作为组合，计算出其乘积，然后，找两者中较为短的一方，慢慢向中间靠拢。

代码如下：

int maxArea(int* height, int heightSize) {int i=0,j=heightSize-1;int area;int result=0;while(i<j){if(height[i]<height[j]){area=height[i]*(j-i);i++;}else{area=height[j]*(j-i);j--;}if(result<area)result=area;}return result;}