LeetCode11: Container With Most Water

Two pointer～

1. Description

• 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.

• 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:

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

Output: 49

2. Two pointer~

参考：Solution

class Solution {
public:
    int maxArea(vector<int>& height) {
        int area = 0;
        int first = 0;
        int end = height.size() - 1; 
        while (first < end){
            area = max(min(height[first], height[end]) * (end - first), area);
            if (height[first] < height[end])
                first++;
            else 
                end--;
        }
        return area;
        
    }
};
           

注意：

1. 木桶盛水的容量取决于最短的那个长度。 这里使用2 pointer法

问题：为什么height[first] < height[end]， 让first ++ ，也就是让first右移？

当左端线段L小于右端线段R时，我们把L右移，这时舍弃的是L与右端其他线段（R-1, R-2, …）组成的木桶，这些木桶是没必要判断的，因为这些木桶的容积肯定都没有L和R组成的木桶容积大。如果高的移动的话，因为一方面高度没有变化（因为取决于最短的木板）而且下标也会变短。