LeetCode-011：Container With Most Water

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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 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

题意：

如图所示，就相当于木桶能装多少水的问题。

思路：

“短板效应”应该很多人都知道吧。一个木桶能装多少水取决于这个木桶的短板有多高。同理，这道题从左右两边开始便利计算容积。设左右指针l、r，根据左右指针的最小高度来计算体积，并比较大小放入最大值maxr变量中。关键：左右指针中移动的是高度最短的那根指针，如果左右高度相等随你移动，爱咋移咋移。退出循环的条件是左右指针相等。

Code：

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