[Swift]LeetCode1064. 不動點 | Fixed Point

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Given an array A of distinct integers sorted in ascending order, return the smallest index i that satisfies A[i] == i.  Return -1 if no such i exists.git

Example 1:github

Input: [-10,-5,0,3,7]
Output: 3 Explanation: For the given array, , thus the output is 3. A[0] = -10, A[1] = -5, A[2] = 0, A[3] = 3

Example 2:數組

Input: [0,2,5,8,17]
Output: 0 Explanation: , thus the output is 0. A[0] = 0

Example 3:微信

Input: [-10,-5,3,4,7,9]
Output: -1 Explanation: There is no such  that , thus the output is -1. iA[i] = i

Note:app

  1. 1 <= A.length < 10^4
  2. -10^9 <= A[i] <= 10^9

給定已經按升序排列、由不一樣整數組成的數組 A,返回知足 A[i] == i 的最小索引 i。若是不存在這樣的 i,返回 -1。 spa

示例 1:code

輸入:[-10,-5,0,3,7]
輸出:3
解釋:
對於給定的數組,,所以輸出爲 3 。
A[0] = -10,A[1] = -5,A[2] = 0,A[3] = 3

示例 2:orm

輸入:[0,2,5,8,17]
輸出:0
示例:
,所以輸出爲 0 。
A[0] = 0

示例 3:htm

輸入:[-10,-5,3,4,7,9]
輸出:-1
解釋: 
不存在這樣的 i 知足 ,所以輸出爲 -1 。A[i] = i

提示:

  1. 1 <= A.length < 10^4
  2. -10^9 <= A[i] <= 10^9

Runtime: 76 ms
Memory Usage: 21.2 MB
 1 class Solution {
 2     func fixedPoint(_ A: [Int]) -> Int {
 3         for i in 0..<A.count
 4         {
 5             if A[i] == i 
 6             {
 7                 return i
 8             }
 9         }
10         return -1
11     }
12 }
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