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Create 11. Container With Most Water.md
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11. Container With Most Water.md
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11. Container With Most Water.md
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# [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)
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# 思路
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题意就是选择两条线,使其组成的容器装的水最多。水的量可以用宽乘高计算。
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由于宽最大就是height两端之间的距离,所以要想在两端之内取得最大值的话只能是高度比较高。
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可以考虑设置两个初始分别为两端的指针left和right代表当前容器,不断跳过高度不够高的height使这两个指针往中间靠。这个过程不断循环即可得到结果。
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时间复杂度O(n),空间复杂度O(1)
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# C++
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``` C++
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class Solution {
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public:
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int maxArea(vector<int>& height) {
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int res = 0;
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int left = 0, right = height.size() - 1;
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while(left < right){
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int h = min(height[left], height[right]); // 当前容器高度
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res = max(res, h * (right - left));
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while(height[left] <= h) left++; // 跳过高度不够高的
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while(height[right] <= h) right--;
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}
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return res;
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}
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};
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```
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