LeetCode 448.Find All Number Disappeared in an Array
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问题
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.
For example:
Input: [4,3,2,7,8,2,3,1]
Output: [2,3]
代码
第一种解法:重新创建一个数组,开始的时候赋值为-1,然后把原数组中的数放在相同的对应的位置上,等于-1的位置就是原数组没有的数字。所用的时间复杂度是O(n),空间复杂度也是O(n)。
具体代码如下:
publicclass Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
List<Integer> res =new ArrayList<Integer>();
if(nums == null || nums.length == 0)
return res;
int n = nums.length;
int[] num = newint[n];
Arrays.fill(num,-1);
for(int i = 0;i < n;i++)
num[nums[i] -1] = nums[i];
for(int i = 0;i < n;i++){
if(num[i] == -1)
res.add(i +1);
}
return res;
}
}
第二种解法:是直接在原数组里改变位置的大小,由于里面的数都在1到n之间,所以把数组里面的数减1,得到一个值,
然后把它当做Index,找到该索引上的值,如果>0,就把该位置上的数变成相反数,最后再统一筛选。
具体代码如下:
publicclass Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
List<Integer> res = new ArrayList<Integer>();
if(nums == null || nums.length == 0)
return res;
int n = nums.length;
for(int i = 0;i < n;i++){
int val = Math.abs(nums[i]) - 1;
if(nums[val] > 0)
nums[val] = -nums[val];
}
for(int i = 0;i < n;i++){
if(nums[i] > 0)
res.add(i + 1);
}
return res;
}
}
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