Leetcode 第30场双周赛

A.转变日期格式
题目链接：https://leetcode-cn.com/problems/reformat-date/

第一眼看到这题就觉得用Python做会很简单，但是我的Python实在是不太行，所以就拿C++硬怼了。

C++：

class Solution {
public:
    string reformatDate(string s) {
        string M[12] = {"Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};
        string MS[12] = {"01", "02", "03", "04", "05", "06", "07", "08", "09", "10", "11", "12"};
        string res = "";
        int len = s.size();
        res = res + s[len - 4] + s[len - 3] + s[len - 2] + s[len - 1] + '-';
        string day = "";
        int index = 0;
        for(int i = 0; i < len; i++) {
            if(s[i] > '9' || s[i] < '0') {
                index = i;
                break;
            } else {
                day = day + s[i];
            }
        }
        if(day.size() == 1) {
            day = '0' + day;
        }
        string month = "";
        for(int i = index + 3; i < len; i++) {
            if(s[i] == ' ') {
                break;
            } else {
                month = month + s[i];
            }
        }
        for(int i = 0; i < 12; i++) {
            if(month == M[i]) {
                month = MS[i];
                break;
            }
        }
        res = res + month + '-' + day;
        return res;
    }
};

之后又学习了Python的解法，真滴快啊：

class Solution:
    def reformatDate(self, s: str) -> str:
        d, m, y = s.split()
        D = int(d[:-2])
        month = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"]
        M = 0
        for i in range(12):
            if month[i] == m:
                M = i+1
        Y = int(y)
        res = "%04d-%02d-%02d" % (Y, M, D)
        return res

B.子数组和排序后的区间和
题目链接：https://leetcode-cn.com/problems/range-sum-of-sorted-subarray-sums/

想了半天有什么优秀算法，然后发现就是暴力。。。

代码如下：

class Solution {
public:
    int rangeSum(vector<int>& a, int n, int left, int right) {
        vector<int> s;
        for (int i = 0; i < n; ++i)
        {
            int tmp = 0;
            for (int j = i; j < n; ++j)
            {
                tmp += a[j];
                s.push_back(tmp);
            }
        }
        sort(s.begin(), s.end());
        int res = 0;
        int MOD = 1e9 + 7;
        for (int i = left-1; i < right; ++i) res = (res+s[i])%MOD;
        return res;
    }
};

C.三次操作后最大值与最小值的最小差
题目链接：https://leetcode-cn.com/problems/minimum-difference-between-largest-and-smallest-value-in-three-moves/

一次一次操作想可能性太多了，全局角度看，就是从排序后的数组两边删掉三个元素，问最后左右两端差值，那么遍历一遍就成了

代码如下：

class Solution {
public:
    int minDifference(vector<int>& nums) {
        int len = nums.size();
        if(len <= 4) {
            return 0;
        }
        sort(nums.begin(), nums.end());
        int res = nums.back() - nums.front();
        for(int i = 0; i <= 3; i++) {
            res = min(res, nums[i + len - 4] - nums[i]);
        }
        return res;
    }
};

D. 石子游戏 IV
题目链接：https://leetcode-cn.com/problems/stone-game-iv/

动态规划

代码如下：

int f[100010];

class Solution {
public:
    bool winnerSquareGame(int n) {
        static int flag = 0;
        if (!flag)
        {
            flag = 1;
            for (int i = 1; i <= 100000; ++i)
            {
                for (int j = 1; j*j <= i; ++j)
                {
                    if (!f[i-j*j])
                    {
                        f[i] = 1;
                        break;
                    }
                }
            }
        }
        return f[n];
    }
};

本文地址：https://blog.csdn.net/weixin_42396397/article/details/107312613