LeetCode [11. Container With Most Water]
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Problem
11. Container With Most Water
Question
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.
思路
直接暴力求解复杂度是O(n^2),如果给出数据太多肯定会超时。
仔细研究可以发现有以下规律,简化计算。
看以下例子:
假设有六段线段:1 2 3 4 5 6用下面的形式简单表现一下两两线段组成的容器从线段1 线段6开始双向遍历 1 2 3 4 5 61 @2 3456存在两种情况1. 线段6比1长或者相等2. 线段6比1短第一种情况下,线段1能组成的最大容器就是和线段6第二种情况下,无法确定最大容器,但有一个规律,线段2、3、4、5就不需要和线段6结合去比较了,具体可看下面例子1 2 3 4 5 6| || | | | | | | | | || | | | | |--------------------------**线段1和6**1 2 3 4 5 6| |_________________________ | || |--------------------------**线段2和6**1 2 3 4 5 6 |____________________ | | | |--------------------------很明显,既然线段6比较短,线段1和线段6组成的容器的宽度必然最大,后面的线段无论是长于6或者短于6,组成的容器必然没有线段1和线段6组成的大(装的水取决于最短的边)。第一种情况下,已经找到线段1的最大容器,那么就让指向线段1的指针右移(不需要和线段2、3、4、5比较了),线段2和线段6比较,以此类推。第二种情况下,指向线段6的指针左移,让线段1和线段5比较。按照以上规则继续进行即可。(第一种情况) 1 2 3 4 5 61 # # # # # @2 @3456(第二种情况) 1 2 3 4 5 61 @ #2 #3 #4 #5 #6 #
综上,代码思路就是,双向遍历,一个从头开始,一个从尾开始,按照以下规则:
- 左边的线段短,左边的位置右移
- 右边的线段短,右边的位置左移
Code
class Solution {public: int maxArea(vector<int>& height) { int left = 0; int right = height.size()-1; int maxVolume = 0; while (left < right) { maxVolume = max(maxVolume, min(height[left], height[right])*(right-left)); if (height[right] >= height[left]) { left++; } else { right--; } } return maxVolume; }};
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