Leetcode题解-11. Container With Most Water

Leetcode题解-11. Container With Most Water

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.

思路

　　一种简单的想法就是用两重遍历，那时间复杂度就是O(n^2)了，比较简单这里就不再赘述了。
　　这里用另外一种方法：用max表示最大容量，开始时用i指向数组头，j指向数组尾，求出i、j之间的容量min{height[i], height[j]}·(j-i)，与max比较，max比当前所求容量小则将当前容量赋值给max。然后倘若height[i]小于height[j]，则i++，反之j- -。重复上述过程直至i等于j，返回max。这里时间复杂度为O(n).
　　证明：假设height[i] < height[j]，container1 =height[i]·(j-i)，则container1 > container2(container2 = height[k]·(k-i)(i < k < j))，所以不必要考虑container2，所以将i++。反之height[i] > height[j]同样的道理，只需要将j- -。

代码

int maxArea(vector<int>& height) {        int i = 0, j = height.size() - 1;        int max = 0;        while(i != j){            int min = 0;            int cur = 0;            if(height[i] <= height[j]){                min = height[i];                cur = min * (j-i);                i++;            }            else{                min = height[j];                cur = min * (j-i);                j--;            }            if(max < cur) max = cur;        }        return max;    }