11. Container With Most Water(容器能包含最多水量)

最新推荐文章于 2024-01-04 13:57:00 发布

原创最新推荐文章于 2024-01-04 13:57:00 发布 · 159 阅读

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本文探讨了如何在给定的垂直线数组中找到能够盛放最多水的两条线段，详细介绍了使用双指针技术解决此问题的高效算法，避免了容器倾斜，并确保至少存在两条线段。

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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

class Solution {
    public int maxArea(int[] height) {
        if( height.length < 2 ){
            return 0;
        }
        //双指针，复杂度O（n）,5ms  91%
        int max=0;
        int i=0;
        int j= height.length-1;
        while(i < j){
            int areaTmp= (j-i) * Math.min(height[i] , height[j]);
            Boolean flag = height[i] > height[j] ? true : false;
            if(max < areaTmp){
                max = areaTmp;                            
            }
            if(flag == true){
                    j--;
                }else{
                    i++;
                }
        }
        
 //        两个for循环，复杂度是O（n2），280+ms，只能超过20%       
//         for(int i=0; i<height.length; i++){
//             for(int j=height.length-1; j>i; j--){       
//                 max= Math.max(max,  (j-i) * Math.min(height[i] , height[j])); 
//             }
//         }
        
        return max;
    }
    

}
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