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.
ref:
http://blog.unieagle.net/2012/09/16/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Acontainer-with-most-water/
思路是从两头到中间扫描,设i,j分别指向height数组的首尾。
那么当前的area是min(height[i],height[j]) * (j-i)。
当height[i] < height[j]的时候,我们把i往后移,否则把j往前移,直到两者相遇。
public int maxArea(int[] height) { int start = 0; int end = height.length - 1; 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; }