Notes

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: ...

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 != s.end(); ++iter) { // failed at here: the logic between conditions is OR not AND if (*iter > '9' || *iter < '0') { break; } p = (*iter - '0') * base; // 7 is the INX_MAX's suffix, remember? if (r > INT_MAX / 10 || (r == INT_MAX / 10 && p > 7)) { return INT_MAX; } // 8 is the INT_MIN's suffix, too. if (r < INT_MIN / 10 || (r == INT_MIN / 10 && p < -8)) { return INT_MIN; } r = r * 10 + p; } return r; } };

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 \(\frac{INT\_MIN}{10}\). ...