題目

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 and n is at least 2.

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:

Input: [1,8,6,2,5,4,8,3,7]
Output:
 
 49

十分鐘嘗試

關鍵是抽象出數學模型，我沒有抽象出來，看了提示後，寫了一下，程式碼如下，二層迴圈

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

答案說有線性時間的解法，我嘗試一下。兩層for迴圈的線性修改就是用兩個指標，程式碼如下：

class Solution {
    public int maxArea(int[] height) {
       int p=0;
       int q=height.length-1;
        int maxArea=0;
       while(p<q){
           maxArea=Math.max(maxArea,Math.min(height[p],height[q])*(q-p));
           p++;
           q--;
       }
        
       return maxArea;
    
    }
}

結果不對，問題出在哪裡？不要兩個指標都移動，這樣不能全部覆蓋。那個低移動哪個。

class Solution {
    public int maxArea(int[] height) {
       int p=0;
       int q=height.length-1;
        int maxArea=0;
       while(p<q){
           maxArea=Math.max(maxArea,Math.min(height[p],height[q])*(q-p));
           if(height[p]<height[q]){
               p++; 
           }
          else{
              q--; 
          }
          
       }
        
       return maxArea;
    
    }
}