LeetCode Daily Question 21

5 月 5 日 每日一题

题目：11. Container With Most Water

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

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Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: 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.

思路

本题是关于双指针和贪心的题目，Greedy 的逻辑为：从可能最大的宽度起始，如果左边小，则 left++，如果右边小则 right–。因为水面高度永远取最小，所以如果移动大的一边，体积必然下降。如果 greedy 后得出的体积比当前的 maxArea 大，则修改当前 maxArea。

代码

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