LeetCode #11 Container With Most Water (M)

[Problem]

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.

 

[Analysis]

这题的主要思路是两边逼近。只需要抓住一个规律：固定较高一边时另一边不断逼近至比当前高度更高的位置，才有可能（不一定）形成更大的面积，即可在O(n)时间完成。

 

[Solution]

public class Solution {
    public int maxArea(int[] height) {
        int max = 0;                
        int head = 0;
        int tail = height.length - 1;
        
        while (head < tail) {
            int currentArea = (tail - head) * (height[head] > height[tail] ? height[tail] : height[head]);
            max = currentArea > max? currentArea : max;
            
            /*
             * if height[head] > height[tail], tracking backwards from tail
             * if height[idx] <= height[tail], the max area between head and idx can't 
             *      exceed the max area between head and tail
             * Reason: in this case for any k < tail, 
             * area(k, head) = (k - head) * min(height[k], height[head])
             * area(tail, head) = (tail - head) * min(height[tail], height[head])
             * area(k, head) < area(tail, head)
             * thus in this case move head until height[k] > height[head] to 
             * get a possibly larger area 
             */
            if (height[head] > height[tail]) {
                int idx = tail; 
                while (idx > head && height[idx] <= height[tail]) {
                    idx--;
                }
                tail = idx;
            } else {
                int idx = head;
                while (idx < tail && height[idx] <= height[head]) {
                    idx++;
                }
                head = idx;
            }
        }
        
        return max;
    }
}