Leetcode Wiggle Sort II

Leetcode Wiggle Sort II，一种比较高效的解法，但是在一些情况下会出现为o(n^2)的时间复杂度。

#include<iostream>#include<vector>#include<algorithm>using namespace std;class Solution {public:    void wiggleSort(vector<int>& nums) {        int sorted = 0;        int cur = 1;        // 1: next one is greater than current sorted last element        // -1: next one is smaller than the current sorted last element        int state = 1;        int pos;        while (sorted < nums.size() - 1) {            if (cur == nums.size()) {                cur = sorted + 1;            }            // if there many elements with the same value between cur and sorted            if (cur != sorted + 1) {                // skip all element with the same value as sorted                while (nums[cur] == nums[sorted]) {                    cur ++;                }                // if cur get the end of the array, let it go back to sorted + 1                if (cur == nums.size()) {                    continue;                }                // if cur does not equals to nums[sorted], we can insert it into                // the sorted sequence.                if (state == 1) {                    // the required next one is greater                    // example 1: 2 5 3 3 4                    // example 2: 2 5 3 3 2                    if (nums[cur] > nums[sorted]) {                        swap(nums[sorted + 1], nums[cur]);                        sorted += 2;                    } else {                        swap(nums[sorted], nums[cur]);                        sorted ++;                        state = -1;                    }                } else {                    // the required next one is smaller                    // example 1: 2 5 2 4 4 5                    // example 2: 2 5 2 4 4 1                    if (nums[cur] > nums[sorted]) {                        swap(nums[sorted], nums[cur]);                        sorted ++;                        state = 1;                    } else {                        swap(nums[sorted + 1], nums[cur]);                        sorted += 2;                    }                }                cur ++;            } else {                // cur = sorted + 1                if (state == 1) {                    // the required next one is greater                    // example 1: 3 5 2 4                    // example 2: 3 5 2 1                    // example 3: 3 5 2 2                    if (nums[cur] > nums[sorted]) {                        sorted ++;                        cur ++;                        state = -1;                    } else if (nums[cur] < nums[sorted]) {                        swap(nums[sorted], nums[cur]);                        sorted ++;                        cur ++;                        state = -1;                    } else {                        pos = search(nums, sorted - 2, state, nums[cur]);                        if (pos != -1) {                            swap(nums[pos], nums[cur]);                        } else {                            // if cannot find the appropriate position, just                            // increase cur.                            cur ++;                        }                    }                } else {                    // example 1: 2 5 4 6 1                    // example 2: 2 5 4 6 7                    // example 3: 2 5 4 6 6                    if (nums[cur] < nums[sorted]) {                        sorted ++;                        cur ++;                        state = 1;                    } else if (nums[cur] > nums[sorted]) {                        swap(nums[sorted], nums[cur]);                        sorted ++;                        cur ++;                        state = 1;                    } else {                        pos = search(nums, sorted - 2, state, nums[cur]);                        if (pos != -1) {                            swap(nums[pos], nums[cur]);                        } else {                            // if cannot find the appropriate position, just                            // increase cur.                            cur ++;                        }                    }                } // end of state = -1            } // end of cur = sorted + 1;            cout<<"current: ";            for (int i = 0; i < nums.size(); i ++) {                cout<<nums[i]<<" ";            }            cout<<endl;        }    }    int search(vector<int>& nums, int pos, int flag, int value) {        // find the appropriate pos for condition: cur = sorte + 1, and        // nums[cur] = num[sorted]        while (pos >= 0) {            if (nums[pos] == value) {                pos -= 2;            } else {                if (flag == 1) {                    if (pos == 0 || value < nums[pos - 1]) {                        return pos;                    } else {                        pos -= 1;                        flag = -1;                    }                } else {                    if (value > nums[pos - 1]) {                        return pos;                    } else {                        pos -= 1;                        flag = 1;                    }                }            }        }        return -1;    }};// Sample input: ./a.out argv1 argv2....int main(int argc, char* argv[]) {    Solution so;    vector<int> vec(atoi(argv[1]), 0);    for (int i = 0; i < atoi(argv[1]); i ++) {        vec[i] = 1 - i % 2;    }    // cout<<"begin: ";    // for (int i = 0; i < vec.size(); i ++) {    //     cout<<vec[i]<<" ";    // }    // cout<<endl;    so.wiggleSort(vec);    cout<<"result: ";    for (int i = 0; i < vec.size(); i ++) {        cout<<vec[i]<<" ";    }    cout<<endl;    return 0;}

测试用例：
代码中给出的一种情况就是o(n^2)的情况，其它的基本没有问题。
./a.out 10
current: 0 1 1 0 1 0 1 0 1 0
current: 0 1 1 0 1 0 1 0 1 0
current: 0 1 0 1 1 0 1 0 1 0
current: 0 1 0 1 1 0 1 0 1 0
current: 0 1 0 1 0 1 1 0 1 0
current: 0 1 0 1 0 1 1 0 1 0
current: 0 1 0 1 0 1 0 1 1 0
current: 0 1 0 1 0 1 0 1 1 0
current: 0 1 0 1 0 1 0 1 0 1
result: 0 1 0 1 0 1 0 1 0 1