【LeetCode with Python】 Container With Most Water
2014-07-06 14:33
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原题页面:https://oj.leetcode.com/problems/container-with-most-water/
题目类型:两指针问题
难度评价:★★★
本文地址:/article/1377510.html
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 linei 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.
用两个指针从数组的左边和右边开始,向中间搜索。依据的理由有两点:
1、因为一开始底边长就已经是最大,两个指针向中间移动的时候,底边长只会变短,因此如果此时面积要变大的话,只能是两条高中的最短者比移动前的最短高更高,否则就无需考察,直接continue到下一次的循环。
2、因此,每次选择移动左指针还是右指针,优先选择当前高度最短的那个,以期寻找到更高的边。如果此时选择移动当前高度最高的那个,就有可能跳过了最优解。
原题页面:https://oj.leetcode.com/problems/container-with-most-water/
题目类型:两指针问题
难度评价:★★★
本文地址:/article/1377510.html
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 linei 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.
用两个指针从数组的左边和右边开始,向中间搜索。依据的理由有两点:
1、因为一开始底边长就已经是最大,两个指针向中间移动的时候,底边长只会变短,因此如果此时面积要变大的话,只能是两条高中的最短者比移动前的最短高更高,否则就无需考察,直接continue到下一次的循环。
2、因此,每次选择移动左指针还是右指针,优先选择当前高度最短的那个,以期寻找到更高的边。如果此时选择移动当前高度最高的那个,就有可能跳过了最优解。
class Solution: # @return an integer def maxArea(self, height): len_height = len(height) if 1 == len_height: return 0 max_area = 0 left_index = 0 right_index = len_height - 1 while left_index < right_index: if height[left_index] < height[right_index]: area = (right_index - left_index) * height[left_index] left_index += 1 else: area = (right_index - left_index) * height[right_index] right_index -= 1 if area > max_area: max_area = area return max_area
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