【LeetCode】11. Container With Most Water 解题小结

 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.

一般的方法就是嵌套两个for循环，不过提交的时候就超时了。那么换个角度考虑，先从最宽的情况，也就是height.size()-1到0这个横坐标之间最大的水容量，比较两者的a_n-1和a₀哪个小一些。然后逐渐往中间靠拢，只有比min(a_n-1，a₀)高的才会被考虑，计算水容量大小，如果比之前大，则更新axis。

class Solution {
public:
    int maxArea(vector<int>& height) {
        int tmpArea = 0;
        int i = 0, j = height.size()-1;
        while (i < j){
            int h = height[i] > height[j] ? height[j]:height[i];
            tmpArea = h * (j - i) > tmpArea ? h * (j - i) : tmpArea;
            while (height[i] <= h && i < j)i++;
            while (height[j] <= h && i < j)j--;
        }
        return tmpArea;
    }
};

转载于:https://www.cnblogs.com/Doctengineer/p/5813127.html