Container With Most Water ——解题笔记
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【题目】
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.
Note: You may not slant the container.
分析:
如果我们遍历两次vector,一定可以得到结果,但是时间复杂度是O(n^2)。
换种思路,两个token,一个放在vector前面,一个放在后面。判断此时的面积是否大于max,然后需要移动两个token,如何移动呢?
如果height[前]小于height[后],那么我们把前面的token后移一位;
如果height[前]大于等于height[后],那么我们把后面的token前移一位;
这样移动的原则是一定要尽量保留较长的边,以保证形成的容积尽可能的大。
代码:
class Solution {public: int maxArea(vector<int>& height) { int area = 0; int len = height.size(); int head = 0; int tail = len - 1; while(head < tail) { int tmp = (tail - head) * min(height[head], height[tail]); area = max(area, tmp); if(height[head] < height[tail]) head++; else tail--; } return area; }};
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