输入：
nums = [5,0,1,2,3,4]
输出：
[4,5,0,1,2,3]
解释：
数组 ans 构建如下：
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]
提示：
- 1 <= nums.length <= 1000
- 0 <= nums[i] < nums.length
- nums 中的元素 互不相同
分析
题目描述既是答案，直接新建一个同样大小的数组，遍历一遍，按顺序赋值即可。
但是这样需要占更多内容空间，而且单纯这样做，根本算不上算法吧？
怎样能做到不占用新的内存空间呢？提示中，每个数的范围都在[0,999]之间，这是个关键。
题解
java
按照题意的直接处理法
class Solution {
public int[] buildArray(int[] nums) {
final int n = nums.length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
ans[i] = nums[nums[i]];
}
return ans;
}
}
不开辟新的空间，直接在原数组处理
class Solution {
public int[] buildArray(int[] nums) {
final int n = nums.length;
final int highValue = 1000;
// 将结果值附加到数组
for (int i = 0; i < n; ++i) {
// 将原值+结果
// 一个值同时存储了原值和结果值
nums[i] += highValue \* (nums[nums[i]] % highValue);
}
// 将原值剔除，仅保留结果
for (int i = 0; i < n; ++i) {
nums[i] /= highValue;
}
return nums;
}
}
c
/\*\*
\* Note: The returned array must be malloced, assume caller calls free().
\*/
int\* buildArray(int\* nums, int numsSize, int\* returnSize){
const int highValue = 1000;
for (int i = 0; i < numsSize; ++i) {
nums[i] += highValue \* (nums[nums[i]] % highValue);
}
for (int i = 0; i < numsSize; ++i) {
nums[i] /= highValue;
}
\*returnSize = numsSize;
return nums;
}
c++
class Solution {
public:
vector<int> buildArray(vector<int>& nums) {
const int highValue = 1000;
int n = nums.size();
for (int i = 0; i < n; ++i) {
nums[i] += highValue \* (nums[nums[i]] % highValue);
}
for (int i = 0; i < n; ++i) {
nums[i] /= highValue;
}
return nums;
}
};
python
from typing import List
class Solution:
def buildArray(self, nums: List[int]) -> List[int]:
highValue = 1000
n = len(nums)
for i in range(n):
nums[i] += highValue \* (nums[nums[i]] % highValue)
for i in range(n):
nums[i] //= highValue
return nums
go
func buildArray(nums []int) []int {
highValue := 1000
n := len(nums)
for i := 0; i < n; i++ {
nums[i] += highValue \* (nums[nums[i]] % highValue)
}
for i := 0; i < n; i++ {
nums[i] /= highValue
}
return nums;
}
rust
impl Solution {
pub fn build\_array(nums: Vec<i32>) -> Vec<i32> {
let mut nums = nums;
let high_value = 1000;
let n = nums.len();
for i in 0..n {
nums[i] += high_value \* (nums[nums[i] as usize] % high_value);
}
for i in 0..n {
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