方法一:
排序之后遍历数组
方法二:
数学方法(较快1ms)
比如[0,1,3]就是1+2+3-数组和
public class Solution {
public int missingNumber(int[] nums) {
int len = nums.length,
result = 0,
i = 0;
Arrays.sort(nums);
for(;i < len;i++) {
if(nums[i] != i) {
result = i;
break;
}
}
return i;
}
}
public class Solution {
public int missingNumber(int[] nums) {
int len = nums.length,
total = 0,
sum = 0,
i = 1;
for(;i <= len;i++) {
total += i;
sum += nums[i - 1];
}
return total - sum;
}
}