You are given a 0-indexed integer array nums
and a target element target
.
A target index is an index i
such that nums[i] == target
.
Return a list of the target indices of nums
after sorting nums
in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.
Example 1:
Input: nums = [1,2,5,2,3], target = 2 Output: [1,2] Explanation: After sorting, nums is [1,2,2,3,5]. The indices where nums[i] == 2 are 1 and 2.
Example 2:
Input: nums = [1,2,5,2,3], target = 3 Output: [3] Explanation: After sorting, nums is [1,2,2,3,5]. The index where nums[i] == 3 is 3.
Example 3:
Input: nums = [1,2,5,2,3], target = 5 Output: [4] Explanation: After sorting, nums is [1,2,2,3,5]. The index where nums[i] == 5 is 4.
Constraints:
1 <= nums.length <= 100
1 <= nums[i], target <= 100
class Solution:
def targetIndices(self, nums: List[int], target: int) -> List[int]:
nums.sort()
return [i for i, num in enumerate(nums) if num == target]
class Solution {
public List<Integer> targetIndices(int[] nums, int target) {
Arrays.sort(nums);
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < nums.length; ++i) {
if (nums[i] == target) {
ans.add(i);
}
}
return ans;
}
}
function targetIndices(nums: number[], target: number): number[] {
nums.sort((a, b) => a - b);
let ans = [];
for (let i = 0; i < nums.length && nums[i] <= target; i++) {
let cur = nums[i];
if (cur == target) {
ans.push(i);
}
}
return ans;
}
class Solution {
public:
vector<int> targetIndices(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
vector<int> ans;
for (int i = 0; i < nums.size(); ++i)
if (nums[i] == target)
ans.push_back(i);
return ans;
}
};
func targetIndices(nums []int, target int) []int {
sort.Ints(nums)
var ans []int
for i, num := range nums {
if num == target {
ans = append(ans, i)
}
}
return ans
}