Notes

tags: Math,backtracking,LeetCode101 Backtracking and stack overflow Intuition: We can abstract all the operations to a Tree, then apply DFS on it. For example: startValue=2, target=3, the tree looks like: /* 2 /\ / \ / \ 1(-1) 4(x2) /\ /\--+ / \ / \ 0(-1) 2(x2) 3(-1) 8(x2) */ class Solution { public: int brokenCalc(int startValue, int target) { unordered_set<int> visited; return backtracking(0, startValue, target, visited); } int backtracking(int count, int val, int target, unordered_set<int> & visited) { if (val == target) { return count; } if (visited.find(val) != visited.end()) { return -1; } int left = -1, right = -1; count++; visited.insert(val); if (val > 0) { left = backtracking(count, val - 1, target, visited); } if (val <= target * 2) { right = backtracking(count, val * 2, target, visited); } visited.erase(val); if (left != -1 && right != -1) { return min(left, right); } else if (left != -1) { return left; } return right; } }; Change target to startValue class Solution { public: int brokenCalc(int startValue, int target) { int r = 0; while (target > startValue) { target = target % 2 > 0 ? target + 1 : target / 2; r++; } return r + startValue - target; } };

tags: String,LeetCode101,Tricky Initialize a string that fills 'a' in it. Then we turn it to the expected string from end to start. The maximal value of each position in the string is 26. If we start from all elements is 'a' in the string. Then the represent value of the string is n. If it’s not equal to k. Then we need turn the last character of string to r = k - n. Two cases must be handled: ...

tags: Math,Hash Table,LeetCode101 class Solution { public: string intToRoman(int num) { vector<string> roman {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}; vector<int> integers {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 }; string r; int times = 0; for (int i = 0; i < integers.size(); ++i) { if (num >= integers[i]) { times = num / integers[i]; num -= times * integers[i]; for (int j = times; j > 0; --j) { r.append(roman[i]); } } } return r; } };

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; } };