leetcode -- Container With Most Water

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

直接BF方法，两个循环，时间复杂度为O(n^2).只能跑过小数据，大数据直接挂了

 1    public int maxArea(int[] height) {
 2         // Start typing your Java solution below
 3         // DO NOT write main() function
 4         int start = 0;
 5         int end = height.length - 1;
 6         int size = height.length;
 7         int max = 0;
 8         for(int i = 0; i < end; i++){
 9             for(int j = end; j > i; j--){
10                 int container = Math.min(height[i], height[j]) * (j - i);
11                 if(container > max)
12                     max  = container;
13             }
14         }
15         return max;
16     }

 参考http://www.mitbbs.com/article_t/JobHunting/32184909.html

本题可以有O(n)时间复杂度的解法，容器容量大小的短板在高度小的那侧，如果height[start] < height[end], 则如果想要是容量变大，唯一做法是使height[start]变大，则start++。height[start] < height[end]时同理使end--

 1 public int maxArea(int[] height) {
 2         int start = 0;
 3         int end = height.length - 1;
 4         int size = height.length;
 5         int max = 0;
 6         while(start < end){
 7             int container = Math.min(height[start], height[end]) * (end - start);
 8             if(max < container){
 9                 max = container;
10             }
11             
12             if(height[start] > height[end]){
13                 end --;
14             } else {
15                 start ++;
16             }
17         }
18         
19         return max;
20     }