Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements that appear twice in this array.
Could you do it without extra space and in O(n) runtime?
Example:
Input:
[4,3,2,7,8,2,3,1] Output:
[2,3]
解法一:
class Solution {
public:
vector<int> findDuplicates(vector<int>& nums) {
vector<int> res;
for (int i = ; i < nums.size(); ++i) {
int idx = abs(nums[i]) - ;
if (nums[idx] < ) res.push_back(idx + );
nums[idx] = -nums[idx];
}
return res;
}
};
下面这种方法是将nums[i]置换到其对应的位置nums[nums[i]-1]上去,比如对于没有重复项的正确的顺序应该是[1, 2, 3, 4, 5, 6, 7, 8],而我们现在却是[4,3,2,7,8,2,3,1],我们需要把数字移动到正确的位置上去,比如第一个4就应该和7先交换个位置,以此类推,最后得到的顺序应该是[1, 2, 3, 4, 3, 2, 7, 8],我们最后在对应位置检验,如果nums[i]和i+1不等,那么我们将nums[i]存入结果res中即可,参见代码如下:
解法二:
class Solution {
public:
vector<int> findDuplicates(vector<int>& nums) {
vector<int> res;
for (int i = ; i < nums.size(); ++i) {
if (nums[i] != nums[nums[i] - ]) {
swap(nums[i], nums[nums[i] - ]);
--i;
}
}
for (int i = ; i < nums.size(); ++i) {
if (nums[i] != i + ) res.push_back(nums[i]);
}
return res;
}
};
下面这种方法是在nums[nums[i]-1]位置累加数组长度n,注意nums[i]-1有可能越界,所以我们需要对n取余,最后要找出现两次的数只需要看nums[i]的值是否大于2n即可,最后遍历完nums[i]数组为[12, 19, 18, 15, 8, 2, 11, 9],我们发现有两个数字19和18大于2n,那么就可以通过i+1来得到正确的结果2和3了,参见代码如下:
解法三:
class Solution {
public:
vector<int> findDuplicates(vector<int>& nums) {
vector<int> res;
int n = nums.size();
for (int i = ; i < n; ++i) {
nums[(nums[i] - ) % n] += n;
}
for (int i = ; i < n; ++i) {
if (nums[i] > * n) res.push_back(i + );
}
return res;
}
};
类似题目:
参考资料:
https://discuss.leetcode.com/topic/64759/very-simple-c-solution
https://discuss.leetcode.com/topic/64735/java-simple-solution/2
https://discuss.leetcode.com/topic/64744/2-pass-o-1-space-solution