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

   public int maxArea(int[] height) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int start = 0;
        int end = height.length - 1;
        int size = height.length;
        int max = 0;
        for(int i = 0; i < end; i++){
            for(int j = end; j > i; j--){
                int container = Math.min(height[i], height[j]) * (j - i);
                if(container > max)
                    max  = container;
            }
        }
        return max;
    }

 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--

public int maxArea(int[] height) {
        int start = 0;
        int end = height.length - 1;
        int size = height.length;
        int max = 0;
        while(start < end){
            int container = Math.min(height[start], height[end]) * (end - start);
            if(max < container){
                max = container;
            }
            
            if(height[start] > height[end]){
                end --;
            } else {
                start ++;
            }
        }
        
        return max;
    }