问题
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?
答案1
建立标记空间,可以是数组可以使HashSet。
public int missingNumber(int[] nums) {
int n = nums.length;
int[] flag = new int[n + 1];
for (int i = 0; i < n; i++) {
flag[nums[i]] = 1;
}
for (int j = 0; j < n + 1; j++) {
if (flag[j] == 0) return j;
}
return -1;
}
答案2
位运算。由于缺失的是0~n中的某一个数,若某数存在,则与该对应标号异或为0。将所有数组中的数与对应标号(0~n)异或,剩余数为缺失数。
public int missingNumber(int[] nums) {
int missing = 0;
for (int i = 0; i < nums.length + 1; i++) missing ^= i;
for (int j = 0; j < nums.length; j++) missing ^= nums[j];
return missing;
}
答案3
将0~n相加,减去数组中所有数,剩余即缺失数。
public int missingNumber(int[] nums) {
//高斯求和公式
int missing = nums.length * (nums.length + 1) / 2;
for (int num : nums) missing -= num;
return missing;
}