Tricky

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....

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: Tricky,LeetCode101,Integer Overflow The key idea is how to detect integer overflow, it’s same to: LeetCode101: 7. Reverse Integer. class Solution { public: int myAtoi(string s) { auto iter = s.begin(); int base = 1, r = 0, p = 0; // Skip whitespace for (; iter != s.end() && *iter == ' '; ++iter) { } // negative or positive if (*iter == '-' || *iter == '+') { if (*iter == '-') { base = -1; } ++iter; } for (; iter !...

tags: Tricky,LeetCode101 Make R = total required rows d = R - 2 2 is contains head line and tail line that not need insert a character between two columns. r = current row offset, which starts from 0. c = current column offset, which starts from 0. We can use a formula to make columns, which is \(c(R+d)+r\). For example, "PAYPALISHIRING", numRows=3: P A H N A P L S I I G Y I R The columns only the head and tail rows is correct should be:...

tags: Tricky,Stack,LeetCode101,Integer Overflow The key is how to detect integer overflow without store to a larger size integer. For this purpose, we could detect integer overflow before carry: The maximal of INT_MAX before carry is \(\frac{INT\_MAX}{10}\). We continue compare pop with the suffix of INT_MAX, 7, if maximal before carry is equal to \(\frac{INT\_MAX}{10}\). The minimal of INT_MIN before carry is \(\frac{INT\_MIN}{10}\) too. We continue compare pop with the suffix of INT_MIN, -8, if minimal before carry is equal to \(\frac{INT\_MIN}{10}\)....

tags: Tricky,LeetCode101 original: 0, 2, 1, 4, 3, 5, 7, 6 max: 0, 2, 2, 4, 4, 5, 7, 7 sorted: 0, 1, 2, 3, 4, 5, 6, 7 index: 0, 1, 2, 3, 4, 5, 6, 7 As shown above, the position of break point is same to the position of max value of chunks. So here: We track chunks’s max value. Break at the position of max value lives in sorted array, which means the index in this case....

tags: Monotonic Stack,LeetCode101,Tricky We travel the numbers in the reverse order: Use a mono-increasing stack to find the largest number(3 in the 132 pattern), the value popped from stack is the second large number(2 in the 132 pattern), if any value less than the second large number, returns true. // Note: // // - subsequence is not contiguous, is i < j < k, not i + 1 = j, j + 1 = k // class Solution { public: bool find132pattern(vector<int>& nums) { int K = INT_MIN; stack<int> mst; // mono-increasing stack for (int i = nums....