题目:
Given n non-negativeintegers a1, a2, ..., an, where eachrepresents a point at coordinate (i, ai). n verticallines 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 thecontainer contains the most water.
Note: You maynot slant the container and n is at least 2.
题意:在二维坐标系中,(i, ai) 表示 从 (i, 0) 到 (i, ai) 的一条线段,任意两条这样的线段和 x 轴组成一个容器,找出能够盛水最多的容器,返回体积;。
思路:用两个指针i,j分别指向数组的头和尾,找出头尾的最小值min(a[i],a[j])并记录体积result1,分别利用判断条件将指针做i++,j--,具体的代码如下,时间复杂度是O(n):
class Solution {
public:
int maxArea(vector<int>& height) {
int i = 0, j = height.size() - 1,result=0;
while (i < j)
{
int h = min(height[j], height[i]);
result = max(result,(j - i)*h);
while(height[i]<=h &&i<j)
i++;
while (height[j] <= h&&i<j)
j--;
}
return result;
}
};