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 = t0Minutes
        for i in range(1, len(timePoints)):
            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

复杂度计算