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Topic description (@ LeetCode)

Given a 32-bit signed integer, you need to reverse the number on each of the integer.

Example 1:

Input: 123 output: 321

Example 2:

Input:-123 output:-321

Example 3:

Input: 120 output: 21

Returns 0 if the inverted integer overflows the 32-bit int range.

Second, the way to solve the problem

Using modular and rounding operations to separate each position, and then accumulate:

123% 10 = 3: take the lowest bit 123 / 10 = 12: remove the lowest bit to check for overflow INT_MAX = 2 ^ 31-1 = 2147483647INT_MIN =-2 ^ 31 =-2147483648 if rec > 2147483647 / 10 = 214748364, then the following rec * 10 must overflow, for example, 2147483650 if rec = 2147483647 / 10 = 214748364 and pop > 7, the same mathematical method class Solution {if the subsequent rec just does not overflow less than 0.

Public:

Int reverse (int x) {

Int rec = 0

Int pop = 0

While (x) {

/ / 1. Take the last place 123% 10 = 3 at a time

Pop = x% 10

/ / 2. Get rid of the last 123 / 10 = 12

X = x / 10

/ / 3. If rec > 2147483647 / 10 = 214748364, subsequent rec * 10 must overflow, such as 214748365 * 10

/ / 4. If rec = 2147483647 / 10 = 214748364 and pop > 7, the subsequent rec just won't overflow.

If ((rec > INT_MAX / 10) | | (rec = = INT_MAX / 10) & & (pop > 7)) return 0

/ / 5. Negative numbers are the same.

If ((rec < INT_MIN / 10) | | (rec = = INT_MIN / 10) & & (pop <-8)) return 0

/ / 6. Cumulative 123 = 12 * 10 + 3

Rec = rec * 10 + pop

}

Return rec

}

}

Complexity Analysis time complexity: O (log (n)) Space complexity: O (1)

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