LeetCode 第30场双周赛

1507. 转变日期格式
字符串处理，呵呵，我的弟弟代码，第一题就写了十分钟

class Solution {
public:
    string reformatDate(string date) {
        vector<string> month = {"","Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};
        char* p = strtok((char*)date.c_str()," ");
        vector<string> v;
        while(p){
            v.push_back(p);
            p = strtok(nullptr," ");
        }
        string ans = "";
        ans += v[2]+"-";
        int m;
        for(int i=1;i<=12;i++){
            if(month[i]==v[1]){
                m = i;
                break;
            }
        }
        if(m<10){
            ans+="0";
        }
        ans += to_string(m)+"-";
        if(v[0][1]>='a' && v[0][1]<='z'){
            ans += "0";
            ans.push_back(v[0][0]);
        }else{
            ans.push_back(v[0][0]);
            ans.push_back(v[0][1]);
        }
        return ans;
    }
};

官方的，巧妙了的运用了stringstream输入流，题给字符串实现了分割。

class Solution {
public:
    string reformatDate(string date) {
        unordered_map<string, string> mp = {
            {"Jan", "01"},
            {"Feb", "02"},
            {"Mar", "03"},
            {"Apr", "04"},
            {"May", "05"},
            {"Jun", "06"},
            {"Jul", "07"},
            {"Aug", "08"},
            {"Sep", "09"},
            {"Oct", "10"},
            {"Nov", "11"},
            {"Dec", "12"}
        };
        stringstream ss(date);//默认以空格分隔
        string year,month,day;
        ss>>day>>month>>year;
        day.pop_back(); //去除最后一个字符
        day.pop_back();
        if(day.size()==1){
            day = "0"+day;
        }
        return year+"-"+mp[month]+"-"+day;
    }
};

1508. 子数组和排序后的区间和
前缀和、暴力
时间复杂度： $O(n^2*log(n))$

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

• k路归并解法
  以nums = [3, 4, 1, 2, 5, 6] 为例 (网友的用例)
  [3, 7, 8, 10, 15, 21]
  [4, 5, 7, 12, 18]
  [1, 3, 8, 14]
  [2, 7, 13]
  [5, 11]
  [6]
  时间复杂度： $O(r*log(n))$
  不过在最坏情况下，仍然是 $O(n^2*log(n))$

class Solution {
public:
    typedef pair<int,int> P;
    int rangeSum(vector<int>& nums, int n, int left, int right) {
        priority_queue<P,vector<P>,greater<P>> pq;
        int cnt = 0 ,ans = 0 , M = 1e9+7;
        for(int i=0;i<n;i++){
            pq.push(make_pair(nums[i],i));
        }
        while(cnt++<right){
            P p = pq.top();
            pq.pop();
            if(cnt>=left){
                ans = (ans+p.first)%M;
            }
            if(p.second+1<n) pq.push(make_pair(p.first+nums[p.second+1],p.second+1));
        }
        return ans;
    }
};

719. 找出第 k 小的距离对
简单

class Solution {
public:
    int minDifference(vector<int>& a) {
        int n = a.size();
        if(n<=4) return 0;
        sort(a.begin(),a.end());
        int ans = a[n-1]-a[0];
        ans = min(ans,a[n-1]-a[3]);
        ans = min(ans,a[n-2]-a[2]);
        ans = min(ans,a[n-3]-a[1]);
        ans = min(ans,a[n-4]-a[0]);
        return ans;
    }
};

1510. 石子游戏 IV
这道DP算是简单的了

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