以Y坐标长度作为木桶边界,以X坐标差为桶底,找出可装多少水。
思路:
前后遍历。
Runtime: 5 ms, faster than 95.28% of Java
class Solution { public int maxArea(int[] height) { int res = 0; int beg = 0; int end = height.length - 1; int temp = 0; while (beg != end) { temp = Math.min(height[beg], height[end]) * (end - beg); res = temp > res ? temp : res; if (height[beg] > height[end]) end--; else beg++; } return res; }