• 268. Missing Number


    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?

    链接: http://leetcode.com/problems/missing-number/

    题解:

    求missing number。首先的想法是先求数列和,再减去数组里的数。这样的话要小心写法,有可能会overflow。我的写法就是会溢出的那种。

    Time Complexity - O(n), Space Complexity - O(1)

    public class Solution {
        public int missingNumber(int[] nums) {
            if(nums == null || nums.length == 0) {
                return 0;
            }
            int sum = (1 + nums.length) * nums.length / 2;
            for(int i : nums) {
                sum -= i;
            }
            return sum;
        }
    }

    也可以用Bit Manipulation,主要就是利用 0 ^ a = a, 以及 a ^ b ^ a = b

    Time Complexity - O(n), Space Complexity - O(1)

    public class Solution {
        public int missingNumber(int[] nums) {
            if(nums == null || nums.length == 0) {
                return 0;
            }
            int res = nums.length;    // nums.length = n
            for(int i = 0; i < nums.length; i++) {
                res ^= i;
                res ^= nums[i];
            }
            return res;
        }
    }

    二刷:

    方法跟一刷一样。 我们首先从把所有0 ~ n的数异或一遍,接着再用这个数把数组中的数异或一遍,这样出现过两次的数都被置零, 剩下的就是 0 ^ missing num = mission num。

    Java:

    public class Solution {
        public int missingNumber(int[] nums) {
            if (nums == null || nums.length == 0) return 0;
            int res = nums.length;
            
            for (int i = 0; i < nums.length; i++) res ^= (i ^ nums[i]);
            
            return res;
        }
    }

    三刷:

    public class Solution {
        public int missingNumber(int[] nums) {
            if (nums == null) return 0;
            int n = nums.length;
            for (int i = 0; i < nums.length; i++) {
                n ^= i;
                n ^= nums[i];
            }
            return n;
        }
    }

    假如输入时排序好的,那么则可以利用index和nums[index]的奇偶性来进行Binary Search

    Reference:

    https://leetcode.com/discuss/53802/c-solution-using-bit-manipulation

    https://leetcode.com/discuss/56174/3-different-ideas-xor-sum-binary-search-java-code

    https://leetcode.com/discuss/53937/simple-c-codes

    https://leetcode.com/discuss/53790/1-lines-ruby-python-java-c

    https://leetcode.com/discuss/53871/java-simplest-solution-o-1-space-o-n-time

    https://leetcode.com/discuss/58647/line-simple-java-bit-manipulate-solution-with-explaination

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  • 原文地址:https://www.cnblogs.com/yrbbest/p/5022828.html
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