第一题 排序数组

Python

C++

Java

第二题 多数元素

Python

C++

Java

第三题 存在重复元素

Python

C++

Java

第四题 最大间距

Python

C++

Java

第五题 按奇偶排序数组

Python

C++

Java

第六题 最小时间差

Python

C++

Java

第七题 三角形的最大周长

Python

C++

Java

第八题 救生艇

Python

C++

Java

总结

C++的sort函数怎么在Leetcode调用？

class Solution {
public:int majorityElement(vector<int>& nums) {sort(nums.begin(), nums.end());return nums[nums.size() / 2];}
};

Java中sort函数调用

class Solution {public int majorityElement(int[] nums) {Arrays.sort(nums);int n = (int)nums.length;return nums[n/2];}
}

C++去重

字符串格式化成数字再转化为分钟

Java

class Solution {public int findMinDifference(List<String> timePoints) {Collections.sort(timePoints);int ans = Integer.MAX_VALUE;int t0Minutes = getMinutes(timePoints.get(0));int preMinutes = t0Minutes;for (int i = 1; i < timePoints.size(); ++i) {int minutes = getMinutes(timePoints.get(i));ans = Math.min(ans, minutes - preMinutes); // 相邻时间的时间差preMinutes = minutes;}ans = Math.min(ans, t0Minutes + 1440 - preMinutes); // 首尾时间的时间差return ans;}public int getMinutes(String t) {return ((t.charAt(0) - '0') * 10 + (t.charAt(1) - '0')) * 60 + (t.charAt(3) - '0') * 10 + (t.charAt(4) - '0');}
}

C++

class Solution {int getMinutes(string &t) {return (int(t[0] - '0') * 10 + int(t[1] - '0')) * 60 + int(t[3] - '0') * 10 + int(t[4] - '0');}public:int findMinDifference(vector<string> &timePoints) {sort(timePoints.begin(), timePoints.end());int ans = INT_MAX;int t0Minutes = getMinutes(timePoints[0]);int preMinutes = t0Minutes;for (int i = 1; i < timePoints.size(); ++i) {int minutes = getMinutes(timePoints[i]);ans = min(ans, minutes - preMinutes); // 相邻时间的时间差preMinutes = minutes;}ans = min(ans, t0Minutes + 1440 - preMinutes); // 首尾时间的时间差return ans;}
};

Python

def getMinutes(t: str) -> int:return ((ord(t[0]) - ord('0')) * 10 + ord(t[1]) - ord('0')) * 60 + (ord(t[3]) - ord('0')) * 10 + ord(t[4]) - ord('0')class Solution:def findMinDifference(self, timePoints: List[str]) -> int:timePoints.sort()ans = float('inf')t0Minutes = getMinutes(timePoints[0])preMinutes = t0Minutesfor i in range(1, len(timePoints)):minutes = getMinutes(timePoints[i])ans = min(ans, minutes - preMinutes)  # 相邻时间的时间差preMinutes = minutesans = min(ans, t0Minutes + 1440 - preMinutes)  # 首尾时间的时间差return ans

https://leetcode-cn.com/problems/minimum-time-difference/solution/zui-xiao-shi-jian-chai-by-leetcode-solut-xolj/

复杂度计算

对n个元素需要做的以n为系数的复杂度就是时间、空间复杂度
比如图上这个，qsort是将n个元素重复n次运行，且每一次需要比较的变得比上一次更少且和剩余的线性相关；故而算 0（xlnx）
然后for只是对于每个n都执行整数次（1）操作，故而算 0（n）