力扣372周赛

力扣第372场周赛

使三个字符串相等

模拟，找到最长前缀，再用每个长度减去最长前缀

class Solution {
public:
    int findMinimumOperations(string s1, string s2, string s3) {
        int a = s1.size() , b = s2.size() , c = s3.size();
        int minl = min(a , min(b , c));
        int maxl = max(a , max(b , c));
        int l = 0 , ans = 0;
        for(; l < minl ; l ++){
            if(s1[l] == s2[l] && (s2[l] == s3[l])){
                continue;
            }else{
                break;
            }
        }
        if(l == 0)return -1;
        ans += (a - l);
        ans += (b - l);
        ans += (c - l);
        return ans;
    }
};
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区分黑球与白球

贪心，找到最近的位置

class Solution {
public:
    long long minimumSteps(string s) {
        long long ans = 0;
        int n = s.size();
        int cnt = 0;
        for(int i = 0 ; i < n ; i ++){
            if(s[i] == '0')cnt ++;
        }
        vector<int>v;
        for(int i = cnt;  i < n ; i ++){
            if(s[i] == '0')v.push_back(i);
        }
        int j = 0;
        for(int i = 0 ; i < cnt ; i ++){
            if(s[i] != '0'){
                ans += (v[j++] - i);
            }
        }
        return ans;
    }
};
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最大异或乘积

单独考虑每一个二进制位，依据数学原理，要让两数尽可能大的同时差尽可能的小

class Solution {
public:
    int maximumXorProduct(long long a, long long b, int n) {
        const int MOD = 1e9 + 7;
        long long p = (a >> n) << n, q = (b >> n) << n;
        for (int i = n - 1; i >= 0; i--) {
            int x = a >> i & 1;
            int y = b >> i & 1;
            if (x == y) p |= 1LL << i, q |= 1LL << i;
            else if (p < q) p |= 1LL << i;
            else q |= 1LL << i;
        }
        p %= MOD;
        q %= MOD;
        return p * q % MOD;
    }
};
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找到 Alice 和 Bob 可以相遇的建筑

后面再补…

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