思路: 模拟

class Solution {
public:int missingInteger(vector<int>& nums) {set<int> s(nums.begin(), nums.end());int acc = nums[0];for (int i = 1; i < nums.size(); i++) {if (nums[i - 1] + 1 != nums[i]) break;acc += nums[i];}while (s.find(acc) != s.end()) {acc ++;}return acc;}
};

思路: 思维题

设 KaTeX parse error: Got function '\prod' with no arguments as superscript at position 5: k ^ \̲p̲r̲o̲d̲_{i=0}^{i=n-1} … 的1个数

class Solution {
public:int minOperations(vector<int>& nums, int k) {int r = 0;for (int v: nums) r ^= v;r = r ^ k;int cnt = 0;while (r > 0) {r = r & (r - 1);cnt++;}return cnt;}
};

思路: BFS

BFS的状态数10^6内

如果x,y值域扩大的话，那就很变难了，需要用dfs来求解

class Solution {
public:int minimumOperationsToMakeEqual(int x, int y) {if (x <= y) return y - x;int inf = 0x3f3f3f3f;vector<int> vec(x + 11 + 1, inf);vec[x] = 0;deque<int> deq;deq.push_back(x);while (!deq.empty()) {int c = deq.front();deq.pop_front();if (c - 1 >= 0 && vec[c - 1] > vec[c] + 1) {vec[c - 1] = vec[c] + 1;deq.push_back(c - 1);} if (c + 1 <= x + 11 && vec[c + 1] > vec[c] + 1) {vec[c + 1] = vec[c] + 1;deq.push_back(c + 1);}if (c % 5 == 0 && vec[c / 5] > vec[c] + 1) {vec[c / 5] = vec[c] + 1;deq.push_back(c / 5);}if (c % 11 == 0 && vec[c / 11] > vec[c] + 1) {vec[c / 11] = vec[c] + 1;deq.push_back(c / 11);}}return vec[y];}
};

思路: 数位DP

using int64 = long long;class Solution {
public:long long numberOfPowerfulInt(long long start, long long finish, int limit, string s) {return solve(finish, limit, s) - solve(start - 1, limit, s);}int64 solve(int64 v, int limit, string s) {vector<int> arr;for (int64 x = v; x > 0; x /= 10) arr.push_back((int)(x % 10));reverse(arr.begin(), arr.end());int n = arr.size(), m = s.length();if (n < s.length()) return 0;if (n == m) {int sv = atoi(s.c_str());return sv <= v ? 1LL : 0LL;}        vector<int64> dp(n, -1);function<int64(int, bool, bool)> dfs = [&](int idx, bool isNum, bool isLimit) {if (idx + m == n) {if (isLimit) {for (int i = idx; i < n; i++) {if (arr[i] < s[i - idx] - '0') return 0LL;else if (arr[i] > s[i - idx] - '0') return 1LL;}}return 1LL;}int64 res = 0;if (!isNum) {res += dfs(idx + 1, false, false);}if (isNum && !isLimit) {if (dp[idx] != -1) return dp[idx];}int s = isNum ? 0 : 1;int e = isLimit ? min(arr[idx], limit) : limit;for (int i = s; i <= e; i++) {res += dfs(idx + 1, true, isLimit && i == arr[idx]);}if (isNum && !isLimit) {dp[idx] = res;}return res;};int64 res = dfs(0, false, true);return res;}};