LeetCode专题：268. Missing Number

Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
Example 1:
Input: [3,0,1]
Output: 2
Example 2:
Input: [9,6,4,2,3,5,7,0,1]
Output: 8
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

给定一个数组，找到缺失的那个数。

解答

import java.util.Arrays;

/**
 Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
 Example 1:
 Input: [3,0,1]
 Output: 2
 Example 2:
 Input: [9,6,4,2,3,5,7,0,1]
 Output: 8
 Note:
 Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
 */
public class MissingNumber {
    public static int missingNumber(int[] nums) {

        /**
         * 方法一：先把数组排序，再遍历数组，相邻两个数相减，如果差值不为1，这两个数中间有缺失
         */
        //排序
        Arrays.sort(nums);

        //对数组首尾进行判断
        if (nums[nums.length-1]!=nums.length) return nums.length;
        if (nums[0]!=0) return 0;

        for (int i=0;i<nums.length-1;i++){
            if (nums[i+1]-nums[i]!=1) return nums[i]+1;
        }

        //方法二；先求和再相减，对nums.length求和
        int result=nums.length*(nums.length+1)/2;
        for (int i=0;i<nums.length;i++){
            result-=nums[i];
        }
         return result;
        //return -1;
    }
    public static void main(String[] args) {
        int[] arr={3,0,1,2,5,6};
        System.out.println(missingNumber(arr));

    }
}