leetcode-287-寻找重复数

 

 方法一： 二分法 O(nlogn) O(1)

class Solution {
    public int findDuplicate(int[] nums) {
        int len = nums.length;
        int left = 1;
        int right = len - 1;
        while(left < right){
            int mid = (left + right) >>> 1;
            int cnt = 0;
            for(int num: nums){
                if(num <= mid){
                    cnt += 1;
                }
            }
            if(cnt > mid){
                right = mid;
            }else{
                left = mid + 1;
            }
        }
        return left;
    }
}

 方法二：位运算 O(nlogn) O(1)

class Solution {
    public int findDuplicate(int[] nums) {
        int n = nums.length,ans = 0;
        int bit_max = 31;
        while(((n - 1) >> bit_max) == 0) {
            bit_max -= 1;
        }
        for (int bit = 0; bit <= bit_max; ++bit){
            int x = 0, y = 0;
            for (int i = 0; i < n; ++i){
                if ((nums[i] & (1 << bit)) != 0) {
                    x += 1;
                }
                if (i >= 1 && ((i & (1 << bit)) != 0)) {
                    y += 1;
                }
            }
            if (x > y){
                ans |= 1 << bit;
            }
        }
        return ans;
    }
}

 方法三：快慢指针 O(n) O(1)

class Solution {
    public int findDuplicate(int[] nums) {
        int slow = 0, fast = 0;
        do {
            slow = nums[slow];
            fast = nums[nums[fast]];
        }while(slow != fast);
        slow = 0;
        while (slow != fast){
            slow = nums[slow];
            fast = nums[fast];
        }
        return slow;
    }
}