Missing Number (E)
题目
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?
题意
给定一个长度为n的数组,从 0-n 这 n+1 个数中选取 n 个数填充入数组中,找到缺失的那一个数。
思路
交换法:第一次遍历数组,将nums[i]与下标为nums[i]的数进行交换,第二次遍历数组,如果遇到 i != nums[i],说明i就是缺失的数。
求和法:先求 0-n 的和,再求数组中数的和,相减就是缺失的数。
异或法:根据相同的数异或为0的性质, 将数组中所有数异或一遍,再将结果与 0-n 异或一遍,相同的数会互相抵消,得到的结果就是缺失的数。
代码实现
Java
交换
class Solution {
public int missingNumber(int[] nums) {
int i = 0;
while (i < nums.length) {
if (nums[i] >= nums.length || nums[i] == i) {
i++;
} else {
swap(nums, i, nums[i]);
}
}
for (int j = 0; j < nums.length; j++) {
if (nums[j] != j) {
return j;
}
}
return nums.length;
}
private void swap(int[] nums, int i, int j) {
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
}
求和
class Solution {
public int missingNumber(int[] nums) {
int len = nums.length;
int sum1 = (len + 1) * len / 2;
int sum2 = 0;
for (int num : nums) {
sum2 += num;
}
return sum1 - sum2;
}
}
异或
class Solution {
public int missingNumber(int[] nums) {
int res = 0;
for (int i = 0; i < nums.length; i++) {
res = res ^ nums[i] ^ (i + 1);
}
return res;
}
}