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  • 260. Single Number III

    Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

    For example:

    Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

    Note:

    1. The order of the result is not important. So in the above example, [5, 3] is also correct.
    2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
    给定一个数字数组,其中只有两个元素只出现一次,而其他所有元素都出现两次。找到只出现一次的两个元素。
    1. /**
    2. * @param {number[]} nums
    3. * @return {number[]}
    4. */
    5. var singleNumber = function(nums) {
    6. let m = {}
    7. for (var i in nums) {
    8. let num = nums[i];
    9. if (m[num]) {
    10. m[num]++;
    11. } else {
    12. m[num] = 1;
    13. }
    14. }
    15. let res = [];
    16. for (var i in m) {
    17. if (res.length == 2) {
    18. break;
    19. }
    20. let num = m[i]
    21. if (num == 1) {
    22. res.push(parseInt(i));
    23. }
    24. }
    25. return res
    26. };






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  • 原文地址:https://www.cnblogs.com/xiejunzhao/p/7652648.html
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