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:

1. You must not modify the array (assume the array is read only).
2. You must use only constant, O(1) extra space.
3. Your runtime complexity should be less than O(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

M1: 在0~n-1范围内，遍历数组。如果nums[abs(nums[i])] >= 0，把它变成小于0；如果< 0，则该数为重复数

时间O(N)，空间O(1)

class Solution {    public int findDuplicate(int[] nums) {        int i;        for(i = 0; i < nums.length; i++) {            if(nums[Math.abs(nums[i])] >= 0)                nums[Math.abs(nums[i])] = - nums[Math.abs(nums[i])];            else                break;        }        return Math.abs(nums[i]);    }}

 

M2: 用binary search，对于由1~n构成的数组，数组长度(nums.length = n + 1)总大于数组里的任意数字，mid = length/2。如果每个数只出现一次，那么必然在mid两边有相同数量的数。如果某一边多了，说明重复的数在这边，可以缩小范围。

从index为1的元素开始统计比较方便。从1~lengh-1，mid是中位数，遍历数组统计<= mid的元素个数cnt，如果没有重复数，<= mid的个数应该大于mid（=mid+1）。在已知存在重复数的情况下，如果<= mid的元素个数<= mid，说明重复数在右侧，l = mid+1；如果>mid，说明在左侧，r = mid。不断重复直到 l >= r，最后的 l 就是答案。

时间O(NlogN)，空间O(1)

class Solution {    public int findDuplicate(int[] nums) {        int l = 1, r = nums.length - 1;        while(l < r) {            int mid = l + (r - l) / 2, cnt = 0;            for(int i : nums) {                if(i <= mid)                    cnt++;            }            if(cnt <= mid)                l = mid + 1;            else                r = mid;        }        return l;    }}