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.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example:
Input: [1,8,6,2,5,4,8,3,7] Output: 49
思路:
这种一行的东西里遍历的题,基本都离不开“哨兵”位,也就是两个“哨兵”分别在最左和最右,然后向里边夹逼。循环的条件一般都是while(left < right),在这个过程中最重要的就是判断哪个哨兵进行移动,在这道题中,我们可以看到“木桶效应”,也就是短板限制了水的多少,所以每次都是哨兵位上板子短的哨兵进行移动,这样就可以跳过很多不可能是最大值的情况,算法效率变高。AC代码如下:
int maxArea(vector<int>& height)
{
int len = height.size();
int l = 0, r = len - 1; // left, right
int max_water = -1;
while(l < r)
{
max_water = max(max_water, (r - l) * min(height[l], height[r]));
if(height[r] < height[l])
--r;
else
++l;
}
return max_water;
}
两个哨兵位相等的时候,++l 和 --r 都可以。所以下边代码也AC:
int maxArea(vector<int>& height)
{
int len = height.size();
int l = 0, r = len - 1; // left, right
int max_water = -1;
while(l < r)
{
max_water = max(max_water, (r - l) * min(height[l], height[r]));
if(height[l] < height[r])
++l;
else
--r;
}
return max_water;
}