You are given an integer array nums.
Return the smallest index i such that the sum of the digits of nums[i] is equal to i.
If no such index exists, return -1.
Example 1:
Input: nums = [1,3,2]
Output: 2
Explanation:
For nums[2] = 2, the sum of digits is 2, which is equal to index i = 2. Thus, the output is 2.
Example 2:
Input: nums = [1,10,11]
Output: 1
Explanation:
For nums[1] = 10, the sum of digits is 1 + 0 = 1, which is equal to index i = 1.
For nums[2] = 11, the sum of digits is 1 + 1 = 2, which is equal to index i = 2.
Since index 1 is the smallest, the output is 1.
Example 3:
Input: nums = [1,2,3]
Output: -1
Explanation:
Since no index satisfies the condition, the output is -1.
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 1000
今天是腦袋放鬆日。
題目問我們給我們一個一維非負整數陣列 nums，
最早使 digitSum 等於索引的是哪個？
這題就是一路做 digitSum 就行了(#。
早停條件可以設定 digitSum 要小於等於 i，
等於是為了考慮後綴 0 的情況。
C++程式碼：
假設 nums 的長度是 N。
計算複雜度：O(N)
-> nums[i] <= 1000，所以最多 O(4*N) ~= O(N)
空間複雜度：O(1)
Github程式碼：
