題目:
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.
分析:若干條線,找兩條線使得其與X軸圍成的面積最大。
如果是暴力的窮搜,肯定可以得到答案,但複雜度達到O(N^2),耗時,leetcode裏面是過不去的。
思路:從數據集的兩頭同時開始。當ai小於aj時,i ++; 反之,j–。直到二者相遇。理由,影響面積的是比較短的線,因此,每次只要改變比較短的線就好了。這個想法的複雜度是O(N)
代碼:
int maxArea(vector<int>& height) {
int len = height.size();
int max_area = 0, area = 0;
int i = 0, j = len - 1;
while(i < j)
{
if(height[i] < height[j])
{
area = (j -i) * height[i];
i ++;
}
else
{
area = (j - i) * height[j];
j --;
}
if (area > max_area)
{
max_area = area;
}
}
return max_area;
}