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.先進行快排,然后在判斷*/ 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; }

轉載于:https://www.cnblogs.com/luozhiyun/p/8367149.html

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