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描述
Given a 0-indexed integer array nums, return the smallest index i of nums such that i mod 10 == nums[i], or -1 if such index does not exist.
x mod y denotes the remainder when x is divided by y.
Example 1:
Input: nums = [0,1,2]
Output: 0
Explanation:
i=0: 0 mod 10 = 0 == nums[0].
i=1: 1 mod 10 = 1 == nums[1].
i=2: 2 mod 10 = 2 == nums[2].
All indices have i mod 10 == nums[i], so we return the smallest index 0.
Example 2:
Input: nums = [4,3,2,1]
Output: 2
Explanation:
i=0: 0 mod 10 = 0 != nums[0].
i=1: 1 mod 10 = 1 != nums[1].
i=2: 2 mod 10 = 2 == nums[2].
i=3: 3 mod 10 = 3 != nums[3].
2 is the only index which has i mod 10 == nums[i].
Example 3:
Input: nums = [1,2,3,4,5,6,7,8,9,0]
Output: -1
Explanation: No index satisfies i mod 10 == nums[i].
Example 4:
Input: nums = [2,1,3,5,2]
Output: 1
Explanation: 1 is the only index with i mod 10 == nums[i].
Note:
1 <= nums.length <= 100
0 <= nums[i] <= 9
解析
根据题意，就是给出来了一个从 0 开始索引的整数列表 nums ，题目要求我们返回最小的索引 i ，使 i mod 10 == nums[i] 成立，如果没有符合题意的索引，那么直接返回 -1 。
x mod y 表示 x 除以 y 的余数。
看起来很难，但是这道题其实很简单，就是遍历 range(len(nums)) 每个索引 i ，如果 i%10 == nums[i] 直接返回 i ，否则遍历结束直接返回 -1 。
解答
class Solution(object):
def smallestEqual(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
for i in range(len(nums)):
if i%10 == nums[i]:
return i
return -1
运行结果
Runtime: 60 ms, faster than 100.00% of Python online submissions for Smallest Index With Equal Value.
Memory Usage: 13.5 MB, less than 100.00% of Python online submissions for Smallest Index With Equal Value.
解析
还可以使用内置函数 next ，直接一行代码就可以搞定。
解答
class Solution(object):
def smallestEqual(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
return next((i for i, x in enumerate(nums) if i%10 == x), -1)
运行结果
Runtime: 52 ms, faster than 100.00% of Python online submissions for Smallest Index With Equal Value.
Memory Usage: 13.4 MB, less than 100.00% of Python online submissions for Smallest Index With Equal Value.
原题链接：leetcode.com/problems/sm…
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