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 the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

Notice that you may not slant the container.

Example 1:

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.

Example 2:

Input: height = [1,1]
Output: 1

Example 3:

Input: height = [4,3,2,1,4]
Output: 16

Example 4:

Input: height = [1,2,1] Output: 2

Constraints:

n == height.length
2 <= n <= 105
0 <= height[i] <= 104

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双重遍历，时间复杂度 O(n²)，曾经用 c++ 写可以通过，现在用 java 会超时。

class Solution {
    public int maxArea(int[] height) {
        int max=0;
        int t;
        for(int i=0;i<height.length-1;++i){
            for(int j=i+1;j<height.length;++j){
                t=Math.min(height[i],height[j])*(j-i);
                if(t>max) max=t;
            }
        }
        return max;
    }
}

两个指针，指向两端，向内移动，每次移动较矮的指针。移动较高的指针，得到的新的容器肯定小于原有的，而移动较矮的，有可能容器比移动前打。最大值肯定出现在每次移动较矮的的步骤中。

class Solution {
    public int maxArea(int[] height) {
        int max=0;
        int t;
        int i=0,j=height.length-1;
        while(i<j){
            t=(j-i)*Math.min(height[i],height[j]);
            max=t>max?t:max;
            if(height[i]<height[j])
                ++i;
            else
                --j;
        }
        return max;
    }
}