leetcode287. Find the Duplicate Number

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本文介绍了一种在不修改原始数组且使用常数级额外空间的情况下查找重复数字的方法。该算法适用于包含n+1个整数的数组，每个整数位于1到n之间，并确保至少存在一个重复数字。

287. Find the Duplicate Number

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.

解法

数组中的元素从1-n，适合做0-1位置的映射，改变对应映射位置的值，检测是否重复。

public class Solution {
    public int findDuplicate(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }

        int len = nums.length;
        for (int i = 0; i < len; i++) {
            int index = (nums[i] - 1) % len;
            if (nums[index] > len) {
                return index + 1;
            } else {
                nums[index] += len;
            }
        }

        return -1;
    }
}