tags: Two Pointers,Tricky,LeetCode101 The key ideas are: We start from two edges and move to the middle with two pointers. Move the pointer to the middle which side is smaller. class Solution { public: int maxArea(vector<int>& height) { int i = 0, j = height.size() - 1; int water = 0; while (i < j) { water = max(water, (j - i) * min(height[i], height[j])); if (height[i] > height[j]) { --j; } else { ++i; } } return water; } };

tags: String,LeetCode101,Stack,Hash Table,Hash Set I have solved this problem years before, LeetCode: 316.Remove Duplicate Letters, but still stuck on it. The key idea is not only about stack, but also required a map to record how many same letters behind current one. Which helps us to decide if drop current letter or not, when the new letter is less than the top of stack, which means smaller in lexicographical order. class Solution { public: string removeDuplicateLetters(string s) { vector<int> countOfLetters(26, 0); vector<bool> pickedLetters(26, false); stack<char> st; for (auto iter = s.begin(); iter != s.end(); ++iter) { countOfLetters[*iter - 'a']++; } for (int i = 0; i < s.size(); ++i) { countOfLetters[s[i] - 'a']--; if (pickedLetters[s[i] - 'a']) { continue; } while (!st.empty() && s[i] < st.top() && countOfLetters[st.top() - 'a'] > 0) { pickedLetters[st.top() - 'a'] = false; st.pop(); } st.push(s[i]); pickedLetters[s[i] - 'a'] = true; } string r; while (!st.empty()) { r.push_back(st.top()); st.pop(); } reverse(r.begin(), r.end()); return r; } };

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; } };