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• If there are no repeating numbers under normal circumstances, then there are only n numbers from 1 to n,
• When there is a repeated number, 1 ~ n has n + 1 numbers.
• So that we can divide the n numbers from the middle number m into two halves,
• If the number in 1-m exceeds m, then the repeated numbers appear in 1 ~ m,
• Conversely, it appears in m + 1 ~ n.

 

• For num, <= the number of elements in the array is count

• When count> num, then 1-

• When count <= num, the number of <= num is not yet

  • There is duplication in num + 1 to n
  • Conversely, there is a duplication from 1 to num-1

 

• The problem becomes, you need to find a number num that satisfies:
  • <= num has more elements than num,
  • <= num-1 The number of elements is less than num.

 

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