27. Remove Element \ 66. Plus One \ 88. Merge Sorted Array

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27. Remove Element

法一:

class Solution {public:    int removeElement(vector<int>& nums, int val) {        for(vector<int>::iterator it = nums.begin();  it != nums.end();)        {            if(*it == val)                it = nums.erase(it);            else                it++;        }        return nums.size();    }};

法二:

class Solution {public:    int removeElement(vector<int>& nums, int val) {        int cnt = 0;        for(int i = 0 ; i < nums.size() ; ++i) {            if(nums[i] == val)                cnt++;            else                nums[i-cnt] = nums[i];        }        return nums.size()-cnt;    }};

66. Plus One

法一:

class Solution {public:    vector<int> plusOne(vector<int> &digits) {        bool carry = true;        for(int i=digits.size()-1; i >= 0 && carry; i--) {            carry = (++digits[i]%=10) == 0;        }        if(carry) {            digits.insert(digits.begin(), 1);        }        return digits;    }};

法二:

class Solution {public:    vector<int> plusOne(vector<int>& digits) {        vector<int> result;        int cf = 0;        if(digits.empty())            return result;        int digits_len = digits.size();        digits[digits_len - 1]++;        int tmp;        for(int it = digits_len - 1; it >= 0; it--)        {            tmp = digits[it] + cf;            digits[it] = tmp%10;            cf = tmp/10;            cout << tmp << " " << cf << endl;        }        if(cf)            result.push_back(cf);        for(int i = 0; i < digits_len; i++)        {            result.push_back(digits[i]);        }        return result;    }};

法一:

class Solution {public:    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {        int ind_1 = 0;        int ind_last = 0;        std::vector<int>::iterator it;        if(n == 0)            return;        if(m == 0)        {            nums1 = nums2;            return;        }        for(int i = 0; i < n; i++)        {            int nums1_len = nums1.size();            for(ind_1 = ind_last; ind_1 < m + i; ind_1++)            {                if(nums2[i] < nums1[ind_1])                {                    it = nums1.begin();                    ind_last = ind_1;                    nums1.insert(it + ind_last, nums2[i]);                    break;                }                else if(ind_1 == m + i - 1)                {                    it = nums1.begin();                    ind_last = ind_1 + 1;                    nums1.insert(it + ind_last, nums2[i]);                    break;                }            }            }        //for(int i = 0; i < n; i++)        //    nums1.pop_back();        nums1.erase(nums1.begin() + m + n, nums1.begin() + m + (n << 1));    }};

法二:

while(n>0)     A[m+n-1] = (m==0||B[n-1] > A[m-1]) ? B[--n] : A[--m];

法三:

class Solution {public:    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {        int i = m - 1, j = n - 1, tar = m + n - 1;        while (j >= 0) {            nums1[tar--] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];        }    }};

参考链接:
http://blog.csdn.net/zhuimengzh/article/details/6841500

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