问题描述：
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.
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.
分析：有n + 1数，其中每个数的范围都是1-n，可以证明这n+1个数中至少有两个数是相同的（鸽巢原理）。假设正好只有两个数是相同的，请找出重复出现的数。
方法一：快速排序，时间复杂度O(nlogn)
public int findDuplicate(int[] nums) {
int n = nums.length; //长度
quickSort(nums,0,n - 1); //快速排序，时间复杂度小于O(n*n)
for(int i = 0; i < n - 1; i++){
if(nums[i] == nums[i + 1] )
return nums[i];
}
return -1;
}
//快速排序
public static void quickSort(int[] nums,int left,int right){
int dp;
if (left < right) {
dp = partition(nums, left, right);
quickSort(nums, left, dp - 1);
quickSort(nums, dp + 1, right);
}
}
//划分
public static int partition(int n[], int left, int right) {
int pivot = n[left];
while (left < right) {
while (left < right && n[right] >= pivot)
right--;
if (left < right)
n[left++] = n[right];
while (left < right && n[left] <= pivot)
left++;
if (left < right)
n[right--] = n[left];
}
n[left] = pivot;
return left;
}
方法二：二分查找
//二分查找 ：若小于mid的数有mid个，则在upperPart查找，否则，在lowerPart查找
public static int findDuplicate(int[] nums){
int n = nums.length - 1; // 总共 n+1 个元素
int low = 1; //数组中可能的最小值
int high = n; //数组中可能的最大值
int mid = 0;
while(low < high){
mid = low + (high - low) / 2; //取中间值
int c = count(mid, nums); //统计小于mid的元素
if(c <= mid){//重复值应该在upperPart
low = mid + 1;
}else { //重复值应该在lowerPart
high = mid - 1;
}
}
return low;
} public static int count(int mid,int[] nums){
int c = 0;
for (int i = 0; i < nums.length; i++) {
if(nums[i] <= mid)
c++;
}
return c;
}