88. 合并两个有序数组

方法一：

class Solution {
public:void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {for(int i = 0; i<n;i++){nums1[m+i] = nums2[i];}sort(nums1.begin(),nums1.end());}
};

方法二：

class Solution {
public:void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {nums1.erase(nums1.begin()+m,nums1.end());nums2.erase(nums2.begin()+n,nums2.end());vector<int> tmp;while (n && m) {if (nums1.front() <= nums2.front()) {tmp.push_back(nums1.front());nums1.erase(nums1.begin());m--;} else {tmp.push_back(nums2.front());nums2.erase(nums2.begin());n--;}}if (!m) {tmp.insert(tmp.end(), nums2.begin(), nums2.end());nums1 = tmp;}if (!n) {tmp.insert(tmp.end(), nums1.begin(), nums1.end());nums1 = tmp;}}
};

方法三：性能最好

class Solution {
public:void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {int p1 = 0, p2 = 0;int sorted[m + n];int cur;while (p1 < m || p2 < n) {if (p1 == m) {cur = nums2[p2++];} else if (p2 == n) {cur = nums1[p1++];} else if (nums1[p1] < nums2[p2]) {cur = nums1[p1++];} else {cur = nums2[p2++];}sorted[p1 + p2 - 1] = cur;}for (int i = 0; i != m + n; ++i) {nums1[i] = sorted[i];}}
};