Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Example 1:
Input: [1,3,4,2,2]
Output: 2
Example 2:
Input: [3,1,3,4,2] Output: 3
Note:
- You must not modify the array (assume the array is read only).
- You must use only constant, O(1) extra space.
- Your runtime complexity should be less than O(n2).
- There is only one duplicate number in the array, but it could be repeated more than once.
方法1:(这个题目比较简单,思路也比较直接)
private int findDuplicateNumber1(int[] nums) {
if (nums.length <= 1) {
return -1;
}
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
if (nums[i] == nums[i + 1]) {
return nums[i];
}
}
return -1;
}
时间复杂度:O(n.logn)
空间复杂度:O(n)
方法2:(利用set集合的形式)
private int findDuplicateNumber2(int[] nums) {
Set<Integer> set = new HashSet<>();
for (int num : nums) {
if (!set.contains(num)) {
set.add(num);
} else {
return num;
}
}
return -1;
}
时间复杂度:O(n)
空间复杂度:O(1)