1、题目
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]
Input:
[0]
Output:
[1]
2、代码实现
public class Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
List<Integer> list = new ArrayList<Integer>();
if (nums == null || nums.length == 0)
return list;
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; ++i) {
map.put(nums[i], 2);
}
for (int i = 1; i <= nums.length; ++i) {
Integer in = map.get(i);
if (in == null) {
list.add(i);
}
}
return list;
}
}
3、总结
当我们需要找到数组集合里面没有包含哪个元素的时候，我们可以采用HashMap来解决这个问题
比如｛1， 2，4， 2， 5｝
还有就是找2个字符串里面第一个不重复元素的问题,我们可以采用HashMap来解决这个问题
s = "leetcode"
return 0.
s = "loveleetcode",
return 2.
或者找字符数组里面唯一一个重复的字符都可以采用HashMap来解决这个问题
