题目

valid-anagram

算法

* o(n)空间复杂度

* o(1)空间复杂度

代码

* O(n)空间复杂度

// O(n) space
class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        vector<int> tmp = nums;
        int n = nums.size(), k = (n + 1) / 2, j = n; 
        sort(tmp.begin(), tmp.end());
        for (int i = 0; i < n; ++i) {
            nums[i] = i & 1 ? tmp[--j] : tmp[--k];
        }
    }
};

* O(1)复杂度

// O(1) space
class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        #define A(i) nums[(1 + 2 * i) % (n | 1)]
        int n = nums.size(), i = 0, j = 0, k = n - 1;
        auto midptr = nums.begin() + n / 2;
        nth_element(nums.begin(), midptr, nums.end());
        int mid = *midptr;
        while (j <= k) {
            if (A(j) > mid) swap(A(i++), A(j++));
            else if (A(j) < mid) swap(A(j), A(k--));
            else ++j;
        }
    }
};