題目
Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
Example:
Input:
[4,3,2,7,8,2,3,1]
Output:
[5,6]
解題思路
將nums[i]置換到其對應的位置nums[nums[i]-1]上去，比如對于沒有缺失項的正確的順序應該是[1, 2, 3, 4, 5, 6, 7, 8]，而我們現在卻是[4,3,2,7,8,2,3,1]，我們需要把數字移動到正確的位置上去，比如第一個4就應該和7先交換個位置，以此類推，最后得到的順序應該是[1, 2, 3, 4, 3, 2, 7, 8]，我們最后在對應位置檢驗，如果nums[i]和i+1不等，那么我們將i+1存入結果res中即可
代碼
findDisappearedNumbers.go
package _448_Find_All_Numbers_Disappeared_in_an_Array
func FindDisappearedNumbers(nums []int) []int {
var ret []int
length := len(nums)
for i := 0; i < length; i++ {
if nums[i] != nums[nums[i] - 1] {
nums[i], nums[nums[i] - 1] = nums[nums[i] - 1], nums[i]
i--
}
}
for i := 0; i < length; i++ {
if (i + 1) != nums[i] {
ret = append(ret, (i+1))
}
}
return ret
}
測試
findDisappearedNumbers_test.go
package _448_Find_All_Numbers_Disappeared_in_an_Array
import "testing"
func sliceEqual(want, ret []int) bool {
len1 := len(want)
len2 := len(ret)
if len1 != len2 {
return false
}
for i := 0; i < len1; i++ {
if want[i] != ret[i] {
return false
}
}
return true
}
func TestFindDisappearedNumbers(t *testing.T) {
input := []int{4,3,2,7,8,2,3,1}
want := []int{5,6}
ret := FindDisappearedNumbers(input)
ok := sliceEqual(want, ret)
if ok {
t.Logf("pass")
} else {
t.Errorf("fail, want %+v, get %+v", want, ret)
}
}
