When the integer number in the array is continuous, calculate the unique repeated integer in the array.

In real life, I guess this problem doesn't make much sense. But as a method, you can also draw a conclusion, that is, to apply mathematics knowledge to programming as much as possible, the algorithm will be perfect. /*************************************** **************************************** **************************************** * ************ 1: problem description: given an array of 101 elements, the elements in the array must belong to [1,100], and each number between [1,100] exists in arr. 2: solution: for analysis, the array contains 101 elements. If each number between [1,100] appears once, it is 100 elements, the remaining element is the unique repeated element. We can traverse arr to get the total number of all elements in arr. Since each number between [1,100] appears once, then we subtract 1 + 2 + ...... + 99 + 100 of the sum, the result is the unique recurring element we need. The time complexity is O (n ). **************************************** **************************************** **************************************** *************/

// Recursive sum of array elements int sum (int a [], int length) {int result = 0; if (length = 1) {result = a [0];} else {result = sum (a, length-1) + a [length-1];} return result;} // use the equals sequence to obtain the sum of 1---100, then subtract it. Int getrepeated (int a []) {int temp1 = (1 + 100) * 100/2; int temp2 = sum (a, 101); return temp2-temp1 ;}