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;}
};

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;}
};

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;}
};

找到 Alice 和 Bob 可以相遇的建筑

离线+最小堆

//代码来自灵茶山艾府
class Solution {
public:vector<int> leftmostBuildingQueries(vector<int> &heights, vector<vector<int>> &queries) {vector<int> ans(queries.size(), -1);//以j为右端点时要处理的第i个的高度，第qi个询问vector<vector<pair<int, int>>> left(heights.size());for (int qi = 0; qi < queries.size(); qi++) {int i = queries[qi][0], j = queries[qi][1];if (i > j) {swap(i, j); // 保证 i <= j}if (i == j || heights[i] < heights[j]) {ans[qi] = j; // i 直接跳到 j} else {left[j].emplace_back(heights[i], qi); // 离线}}//最小堆priority_queue<pair<int, int>, vector<pair<int, int>>, greater<>> pq;for (int i = 0; i < heights.size(); i++) { // 从小到大枚举下标 i//右端点为i的询问入堆for (auto &p: left[i]) {pq.emplace(p); // 后面再回答}//堆中小于的即为答案while (!pq.empty() && pq.top().first < heights[i]) {ans[pq.top().second] = i; // 可以跳到 i（此时 i 是最小的）pq.pop();} }return ans;}
};

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