11. Container With Most Water
Problem
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
Solution (贪心)
1、C 8 ms, faster than 65.90% 8 MB, less than 44.59%
int maxArea(int* height, int heightSize)
{
int left = 0, right = heightSize - 1;
int maxArea = 0, currentArea, shortline;
while (left < right) {
shortline = (*(height + left)) < (*(height + right)) ? (*(height + left)) : (*(height + right));
currentArea = shortline * (right - left);
maxArea = maxArea > currentArea ? maxArea : currentArea;
if (*(height + left) <= *(height + right)) {
while (left < right && *(height + left) <= shortline) {
left++;
}
} else {
while (left < right && *(height + right) <= shortline) {
right--;
}
}
}
return maxArea;
}
2、python 44 ms, faster than 49.21% 11.9 MB, less than 5.21%
class Solution(object):
def maxArea(self, height):
"""
:type height: List[int]
:rtype: int
"""
left = 0
right = len(height) - 1
shortline, maxArea = 0, 0
while left < right:
shortline = height[left] if height[left] < height[right] else height[right]
currentArea = shortline * (right - left)
maxArea = maxArea if maxArea > currentArea else currentArea
if height[left] < height[right]:
while left < right and height[left] <= shortline:
left += 1
else:
while left < right and height[right] <= shortline:
right -= 1
return maxArea