LeetCode - Algorithms
  • Introduction
  • 1. Two Sum
  • 2. Add Two Numbers
  • 3. Longest Substring Without Repeating Characters
  • 4. Median of Two Sorted Arrays
  • 5. Longest Palindromic Substring
  • 6. ZigZag Conversion
  • 7. Reverse Integer
  • 8. String to Integer (atoi)
  • 9. Palindrome Number
  • 10. Regular Expression Matching
  • 11. Container with Most Water
  • 12. Integer to Roman
  • 13. Roman to Integer
  • 14. Longest Common Prefix
  • 15. 3Sum
  • 16. 3Sum Closest
  • 17. Letter Combinations of a Phone Number
  • 18. 4Sum
  • 19. Remove Nth Node From End of List
  • 20. Valid Parentheses
  • 21. Merge Two Sorted Lists
  • 22. Generate Parentheses
  • 23. Merge k Sorted Lists
  • 24. Swap Nodes in Pairs
  • 25. Reverse Nodes in k-Group
  • 26. Remove Duplicates from Sorted Array
  • 27. Remove Element
  • 28. Implement strStr()
  • 29. Divide Two Integers
  • 30. Substring with Concatenation of All Words
  • 31. Next Permutation
  • 32. Longest Valid Parentheses
  • 33. Search in Rotated Sorted Array
  • 34. Search for a Range
  • 35. Search Insert Position
  • 36. Valid Sudoku
  • 37. Sudoku Solver
  • 38. Count and Say
  • 39. Combination Sum
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7. Reverse Integer

Problem

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:

Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Related Topics:

Math

Analysis

重点在于溢出处理,可以参考 Integer.parseInt() 方法,核心思想是:

  • 统一采用负数来处理。

  • 提前缩小区间范围,避免溢出。

Code

class Solution {

    fun reverse(x: Int): Int {

        var xi = x
        var result = 0
        var negative = false
        var limit = -Int.MAX_VALUE
        val multmin: Int
        var digit: Int

        if (xi < 0) {
            negative = true
            limit = Int.MIN_VALUE
        }

        multmin = limit / 10
        while (xi != 0) {

            digit = if (negative) -(xi % 10) else xi % 10
            xi /= 10

            if (result < multmin) {
                return 0
            }
            result *= 10

            if (result < limit + digit) {
                return 0
            }
            result -= digit
        }

        return if (negative) result else -result
    }
}
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