Task

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. leetcode11 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

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Input: [1,8,6,2,5,4,8,3,7]
Output: 49

Solution

  • 使用双指针l和r指向两个端点的木板，向中间移动
  • 若左侧木板更高，应继续尝试更高木板和其他木板的组合，即将右侧木板向左移动
  • 若右侧木板更高，应继续尝试更高木板和其他木板的组合，即将左侧木板向右移动
  • 面积公式是min{height[l],height[r]}*(r-l)
  • 时间复杂度：只遍历一次，O(n)
  • 空间复杂度：只用了指针和返回值，O(1)
  • 实现：
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// Runtime: 28 ms, faster than 53.05% of C++ online submissions for Container With Most Water.
// Memory Usage: 14.4 MB, less than 5.15% of C++ online submissions for Container With Most Water.
class Solution {
public:
    int maxArea(vector<int>& height) {
        int l=0,r=height.size()-1;
        int ans=0;
        //双指针遍历
        while(l!=r){
            //min(height[l],height[r])*(r-l)是面积，求max是保留最大面积
            ans=max(ans,min(height[l],height[r])*(r-l));、
            //左侧更高时，应尝试左侧和其他木板的组合，故右侧木板向左推进
            if(height[l]>height[r])
                --r;
            else
                ++l;
        }
        return ans;
    }
};