[leetcode]11. Container With Most Water存水最多的容器
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
题意:
给定一堆高矮不一的平行的墙,选择其中两堵作为两壁,问最多能存多少水。
Solution1:
Two Pointers: set pointer left at index 0; set pointer right at index len - 1
We calculate area base on width (gap of left and right index ) and height(smaller height matters because of "Short-board effect" )
In order to find max area, we move the pointer whichever representing the smaller height
code:
/*
Time complexity : O(n). We traverse the whole give array
Space complexity : O(1). We only used constant extra space.
*/ class Solution {
public int maxArea(int[] height) {
int left = 0;
int right = height.length - 1;
int area = Integer.MIN_VALUE;
while(left < right){
area = Math.max(area, Math.min(height[left], height[right]) * (right - left));
if(height[left] < height[right]){
left ++;
}else{
right --;
}
}
return area;
}
}
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