comments | difficulty | edit_url | rating | source | tags | ||
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true |
Easy |
1160 |
Weekly Contest 248 Q1 |
|
Given a zero-based permutation nums
(0-indexed), build an array ans
of the same length where ans[i] = nums[nums[i]]
for each 0 <= i < nums.length
and return it.
A zero-based permutation nums
is an array of distinct integers from 0
to nums.length - 1
(inclusive).
Example 1:
Input: nums = [0,2,1,5,3,4] Output: [0,1,2,4,5,3] Explanation: The array ans is built as follows: ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]] = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]] = [0,1,2,4,5,3]
Example 2:
Input: nums = [5,0,1,2,3,4] Output: [4,5,0,1,2,3] Explanation: The array ans is built as follows: ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]] = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]] = [4,5,0,1,2,3]
Constraints:
1 <= nums.length <= 1000
0 <= nums[i] < nums.length
- The elements in
nums
are distinct.
Follow-up: Can you solve it without using an extra space (i.e., O(1)
memory)?
class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
return [nums[num] for num in nums]
class Solution {
public int[] buildArray(int[] nums) {
int[] ans = new int[nums.length];
for (int i = 0; i < nums.length; ++i) {
ans[i] = nums[nums[i]];
}
return ans;
}
}
class Solution {
public:
vector<int> buildArray(vector<int>& nums) {
vector<int> ans;
for (int& num : nums) {
ans.push_back(nums[num]);
}
return ans;
}
};
func buildArray(nums []int) []int {
ans := make([]int, len(nums))
for i, num := range nums {
ans[i] = nums[num]
}
return ans
}
function buildArray(nums: number[]): number[] {
return nums.map(v => nums[v]);
}
impl Solution {
pub fn build_array(nums: Vec<i32>) -> Vec<i32> {
nums.iter().map(|&v| nums[v as usize]).collect()
}
}
/**
* @param {number[]} nums
* @return {number[]}
*/
var buildArray = function (nums) {
let ans = [];
for (let i = 0; i < nums.length; ++i) {
ans[i] = nums[nums[i]];
}
return ans;
};
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* buildArray(int* nums, int numsSize, int* returnSize) {
int* ans = malloc(sizeof(int) * numsSize);
for (int i = 0; i < numsSize; i++) {
ans[i] = nums[nums[i]];
}
*returnSize = numsSize;
return ans;
}