牛客网/剑指OFFER/（3）数组中重复的数字_*duplication = numbers[i];运行时出错-优快云博客

本文介绍了一种在整数数组中查找重复数字的算法。该算法首先验证输入的有效性，然后通过排序或原地交换元素的方法来定位重复的数字。文章提供了两种不同的实现方式，并详细解释了每一步的操作。

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class Solution {
public:
	// Parameters:
	//        numbers:     an array of integers
	//        length:      the length of array numbers
	//        duplication: (Output) the duplicated number in the array number
	// Return value:       true if the input is valid, and there are some duplications in the array number
	//                     otherwise false
	bool duplicate(int numbers[], int length, int* duplication) {
		int x;

		if (numbers == nullptr || length <= 0)
			return false;
		for (int i = 0; i < length; ++i) {
			if (numbers[i] > length - 1 || numbers[i] < 0)
				return false;
		}

		sort(numbers, numbers + length);//升序排列
		for (int j = 0; j < length; ++j) {
			if (numbers[j] == numbers[j + 1]) {
				*duplication = numbers[j];
				return true;
			}
		}
		return false;
	}
};

class Solution {
public:
	// Parameters:
	//        numbers:     an array of integers
	//        length:      the length of array numbers
	//        duplication: (Output) the duplicated number in the array number
	// Return value:       true if the input is valid, and there are some duplications in the array number
	//                     otherwise false
	bool duplicate(int numbers[], int length, int* duplication) {
		int x;

		if (numbers == nullptr || length <= 0)
			return false;
		for (int i = 0; i < length; ++i) {
			if (numbers[i] > length - 1 || numbers[i] < 0)
				return false;
		}

		for (int i = 0; i < length; ++i) {
			while (numbers[i] != i) {
				if (numbers[i] == numbers[numbers[i]]) {
					*duplication = numbers[i];
					return true;
				}
				else
				{
					int temp = numbers[i];
					numbers[i] = numbers[numbers[i]];
					numbers[temp] = temp;
				}

			}
		}	
		return false;
	}
};