头图

JZ-050 - Repeating numbers in array

雄狮虎豹
中文

Repeated numbers in an array

Topic description

All numbers in an array of length n are in the range 0 to n-1. Some numbers in the array are repeated, but I don't know how many numbers are repeated. Also don't know how many times each number is repeated. Please find in the array

  • The first repeated number. For example, if the input array {2,3,1,0,2,5,3} of length 7, then the corresponding output is the first repeated number 2.
  • Return description:
  • The function returns true if there are duplicate numbers in the array, false otherwise.
  • If there are duplicate numbers in the array, put the duplicate numbers in the parameter duplication[0]. (ps:duplication has been initialized and can be used directly by assignment.)

Topic link : Repeated numbers

code

/**
 * 标题:数组中重复的数字
 * 题目描述
 * 在一个长度为n的数组里的所有数字都在0到n-1的范围内。 数组中某些数字是重复的,但不知道有几个数字是重复的。也不知道每个数字重复几次。请找出数组中
 * 第一个重复的数字。 例如,如果输入长度为7的数组{2,3,1,0,2,5,3},那么对应的输出是第一个重复的数字2。
 * 返回描述:
 * 如果数组中有重复的数字,函数返回true,否则返回false。
 * 如果数组中有重复的数字,把重复的数字放到参数duplication[0]中。(ps:duplication已经初始化,可以直接赋值使用。)
 * <p>
 * 题目链接
 * https://www.nowcoder.com/practice/623a5ac0ea5b4e5f95552655361ae0a8?tpId=13&&tqId=11203&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking
 */
public class Jz50 {

    /**
     * 暴力破解
     *
     * @param numbers
     * @param length
     * @param duplication
     * @return
     */
    public static boolean duplicate(int[] numbers, int length, int[] duplication) {
        if (length <= 1) {
            return false;
        }

        for (int i = 0; i < length - 1; i++) {
            for (int j = i + 1; j < length; j++) {
                if (numbers[j] == numbers[i]) {
                    duplication[0] = numbers[i];
                    return true;
                }
            }
        }
        return false;
    }

    public static boolean duplicate1(int[] nums, int length, int[] duplication) {
        if (nums == null || length <= 0) {
            return false;
        }
        for (int i = 0; i < length; i++) {
            while (nums[i] != i) {
                if (nums[i] == nums[nums[i]]) {
                    duplication[0] = nums[i];
                    return true;
                }
                swap(nums, i, nums[i]);
            }
        }
        return false;
    }

    public static void swap(int[] nums, int i, int j) {
        int t = nums[i];
        nums[i] = nums[j];
        nums[j] = t;
    }

    public static void main(String[] args) {
        int[] numbers = {2, 3, 1, 0, 2, 5, 2};
        int[] duplication = new int[1];
        int[] duplication2 = new int[1];
        int[] duplication11 = new int[1];
        boolean duplicate = duplicate(numbers, 7, duplication);
        System.out.println(duplicate);
        System.out.println(duplication[0]);

        boolean duplicate1 = duplicate1(numbers, 7, duplication2);
        System.out.println(duplicate1);
        System.out.println(duplication2[0]);
    }
}
[Daily Message] When you are unhappy, you can eat a piece of candy and tell yourself that life is still sweet, come on.
阅读 628

LeetCodet题解
技术学习

玉树临风,仙姿佚貌!

1.5k 声望
6.5k 粉丝
0 条评论

玉树临风,仙姿佚貌!

1.5k 声望
6.5k 粉丝
文章目录
宣传栏