268. 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

我的版本:

package leetCode;

import java.util.Arrays;

/**
 * Created by luoluo on 2018/1/27.
 	先进行快排,然后在判断
 */
public class MissingNumber {

    public static void main(String[] args) {
        int[] a = {3,0,1};
        missingNumber(a);
        System.out.println(Arrays.toString(a));
    }

    public static int missingNumber(int[] nums) {
        int n = nums.length;

        sort(nums, 0, n-1);

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

    public static void sort(int[] nums ,int bg,int end){
        if (bg >= end) {
            return;
        }
        int partition = partition(nums, bg, end);
        sort(nums, bg, partition - 1);
        sort(nums, partition + 1, end);
    }

    public static int partition(int[] nums,int bg, int end){
        int l = nums[bg];

        int j = bg;
        for (int i = bg+1; i <= end; i++) {
            if(nums[i]>l){
                j++;
                swap(nums, i, j);
            }
        }
        swap(nums, j, bg);
        return j;
    }

    public static void swap(int[] nums ,int a ,int b) {
        int temp = nums[a];
        nums[a] = nums[b];
        nums[b] = temp;
    }
}

大神的版本:

public static int missingNumber(int[] nums) {
    int sum = nums.length;
    for (int i = 0; i < nums.length; i++)
        sum += i - nums[i];
    return sum;
}