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:

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.

思考路线：

1） 排序后遍历：不符合条件1的要求，数组不可改；

2） map：不符合条件2的要求，O(n)的空间复杂度。

思路：针对2，想起了First Missing Number的解法，数组既保存原有数组信息，又保存map信息。直接遍历数组nums，

遍历到元素i,将nums[i]置为-1*nums[i],此时nums[i]为负；若下次又遍历到i,即可检查到nums[i]为负，即可知道数字i为Dumplicate数。

误区：写的过程中，进入误区了，只要遍历遇到i，则将nums[i]置为-1*nums[i]，遍历完后再次遍历，若是有负数，即对应的下标为

Dumplicate数。这样考虑是因为以为只会重复两次，但是题目并没有这么说，即如果存在多次（比如偶数次），最后还是可能转为正数。

代码：

public class Solution {
    public int findDuplicate(int[] nums) {
        int index, length = nums.length, result = 0;
        for(int i = 0; i < length; i++) {
        	index = nums[i];
        	if(index < 0) {
        		index *= -1;
        	}
        	if(nums[index] < 0) {
        	    result = index;
        	    break;
        	} else {
        		nums[index] *= -1;
        	}
        }
        return result;
    }
}

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