[LeetCode] 961. N-Repeated Element in Size 2N Array

題目

In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.

Return the element repeated N times.

Note:

• 4 <= A.length <= 10000
• 0 <= A[i] < 10000
• A.length is even

直覺解

構想：

因為陣列中的元素可能是 0，所以不能用正負號來紀錄數字是否出現過，也不能把陣列轉換成 linked list。

從 nums 的第一個元素開始檢查，把出現過的元素記錄下來。若發現有個數字已經出現過，就 return 該數字

javascript 實作：

const repeatedNTimes = function(A) {

   const seen = {}

   let i = 0

   while(i < A.length) {
       if (!seen[A[i]]) {

           seen[A[i]] = true

       } else {

            return A[i]

       }

        i++
   }

}

效能：

Runtime: 72 ms, faster than 98.97% of JavaScript online submissions for N-Repeated Element in Size 2N Array.
Memory Usage: 41.8 MB, less than 93.08% of JavaScript online submissions for N-Repeated Element in Size 2N Array.

犯過的低級錯誤：

• 把 seen[A[i]] 寫成 seen.[A[i]]