Notes

tags: String,LeetCode101,Hash Set,Hash Table,Stack The key idea is similar to LeetCode101: 316. Remove Duplicate Letters, we use a HashMap to track how many letters which is same in the string. Then we use a HashSet to store appeared letters. When there is no more letters appeared in the HashSet, it’s time to partition. class Solution { public: vector<int> partitionLabels(string s) { unordered_map<char, int> cntOfLetters; unordered_set<char> appearedLetters; vector<int> r; int count = 0; for (auto iter = s.begin(); iter != s.end(); ++iter) { cntOfLetters[*iter]++; } for (auto iter = s.begin(); iter != s.end(); ++iter) { appearedLetters.insert(*iter); cntOfLetters[*iter]--; count++; // Append current count to result if there is no appeared letters // behind. bool behind = false; for (auto si = appearedLetters.begin(); si != appearedLetters.end(); ++si) { if (cntOfLetters[*si]){ behind = true; break; } } if (!behind) { r.push_back(count); count = 0; } } if (count > 1) { r.push_back(count); } return r; } };

tags: Algorithm

tags: LeetCode101,Math The key idea is: To use \(log10(10^n) = n\) to get how many digits in the number. Then we need iterate \(n + 1\) times to compare each side. The digit in the left is \(\frac{x}{10^{n-i}} \mod 10\). The digit in the right is \(\frac{x}{10^i} \mod 10\). class Solution { public: bool isPalindrome(int x) { if (x < 0) { return false; } // failed at here if (x < 10) { return true; } int n = log10(x); int ld = pow(10, n); // left div int rd = 1; // right div for (int i = 0; i < (n + 1) / 2; i++) { // left right if (x / ld % 10 != x / rd % 10) { return false; } ld /= 10; rd *= 10; } return true; } };

tags: Algorithm

tags: C/C++ In some problems, we need to detect is our result overflow in a 32-bit integer. The key ideas is check our value before it becomes bigger. For example: // INT_MAX 2147483647 // INT_MIN -2147483648 // INT_MAX's suffix is 7 if (res > INT_MAX / 10 || (res == INT_MAX / 10 && pop > 7)) { return 0; } // INT_MIN's suffix is -8 if (res < INT_MIN / 10 || (res == INT_MIN / 10 && pop < -8)) { return 0; } res = res * 10 + pop; Our final result need a 10 times current value and plus a value, then we check: ...