A

class Solution {
public:
    vector<int> sortEvenOdd(vector<int>& nums) {
        vector<int> a, b, v;
        for (int i = 0; i < nums.size(); i ++ )
            if (i % 2 == 0) a.push_back(nums[i]);
            else b.push_back(nums[i]);
        sort(a.begin(), a.end());
        sort(b.begin(), b.end());
        reverse(b.begin(), b.end());
        for (int i = 0; i < min(a.size(), b.size()); i ++ )
             v.push_back(a[i]), v.push_back(b[i]);
        if (a.size() > b.size())
             v.push_back(a[a.size() - 1]);
        return v;
    }
};

B

class Solution {
public:
    int cnt[10];
    long long smallestNumber(long long num) {
        if (num >= 0) {
            while (num) {
                cnt[num % 10] ++;
                num /= 10;
            }
            long long res = 0;
            for (int i = 1; i <= 9; i ++ ) {
                if (cnt[i]) {
                    res = i;
                    cnt[i] --;
                    break;
                }
            }
            while (cnt[0] -- )
                res = res * 10;
            for (int i = 1; i <= 9; i ++ ) {
                while (cnt[i]) {
                    res = res * 10 + i;
                    cnt[i] --;
                }
            }
            return res;
        }
        else {
            num *= -1;
            long long res = 0;
            while (num) {
                cnt[num % 10] ++;
                num /= 10;
            }
            for (int i = 9; i >= 0; i -- ) {
                while (cnt[i]) {
                    res = res * 10 + i;
                    cnt[i] --;
                }
            }
            return res * -1;
        }
    }
};

C

class Bitset {
public:
    int len, cnt = 0, tot = 0;
    int a[100010] = {0};
    Bitset(int size) {
        len = size;
    }
    
    void fix(int idx) {
        if (a[idx] == 0 && tot % 2 == 0)    cnt ++, a[idx] = 1;
        if (a[idx] == 1 && tot % 2 == 1)    cnt ++, a[idx] = 0;
    }
    
    void unfix(int idx) {
        if (a[idx] == 1 && tot % 2 == 0)    cnt --, a[idx] = 0;
        if (a[idx] == 0 && tot % 2 == 1)    cnt --, a[idx] = 1;
    }
    
    void flip() {
        cnt = len - cnt;
        tot ++;
    }
    
    bool all() {
        if (cnt == len) return true;
        else    return false;
    }
    
    bool one() {
        if (cnt >= 1)   return true;
        else return false;
    }
    
    int count() {
        return cnt;
    }
    
    string toString() {
        string s;
        for (int i = 0; i < len; i ++ ) {
            if (tot & 1) {
                if (a[i])   s += "0";
                else s += "1";
            }
            else {
                if (a[i])   s += "1";
                else s += "0";
            }
        }
        return s;
    }
};

D
从两个方向DP，分别记作\(f1,f2\)，那么答案就为\(min\{f1[i]+f2[i + 1]\}\)

class Solution {
public:
    char s[200010];
    int f1[200010], f2[200010];
    int minimumTime(string ss) {
        int n = ss.size();
        memset(f1, 0x3f, sizeof f1);
        memset(f2, 0x3f, sizeof f2);
        for (int i = 0; i < ss.size(); i ++ )   s[i + 1] = ss[i];
        f1[0] = f2[n + 1] = 0;
        for (int i = 1; i <= n; i ++ ) {
            f1[i] = f1[i - 1];
            if (s[i] == '1') {
                f1[i] = min(i, f1[i] + 2);
            }
        }
        for (int i = n; i >= 1; i -- ) {
            f2[i] = f2[i + 1];
            if (s[i] == '1') {
                f2[i] = min(n - i + 1, f2[i + 1] + 2);
            }
        }
        int res = 0x3f3f3f3f;
        for (int i = 0; i <= n; i ++ ) {
            res = min(res, f1[i] + f2[i + 1]);
        }
        return res;
    }
};