【LeetCode】268. Missing Number
2016-02-25 22:48
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For example,
Given nums = [0, 1, 3] return 2.
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
遍历数组，所有数相加和 减去sum就是missing的数
O(n)
题目
Given an array containing n distinct numbers taken from 0, 1, 2, …, n, find the one that is missing from the array.For example,
Given nums = [0, 1, 3] return 2.
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
public class Solution {
public int missingNumber(int[] nums) {
int n = nums.length;
int sum = ((1 + n) * n )/ 2;
int sum1 = 0;
for(int i = 0 ; i < n ;i++)
{
sum1 += nums[i];
}
int ans = sum - sum1;
return ans;
}
}思路
n项和公式 sum = (n + 1) * n / 2遍历数组，所有数相加和 减去sum就是missing的数
O(n)
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