64 lines
1.9 KiB
C++
64 lines
1.9 KiB
C++
class Solution {
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public:
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vector<int> mergeArrays(vector<int>& v1, vector<int>& v2){
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int size1 = v1.size();
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int size2 = v2.size();
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vector<int> v(size1+size2, 0);
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int index1 = 0;
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int index2 = 0;
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int index = 0;
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// scan and compare, whoever is smaller goes to v.
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while(index1<size1 && index2<size2){
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if(v1[index1]<v2[index2]){
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v[index] = v1[index1];
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index1 = index1 + 1;
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}else{
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v[index] = v2[index2];
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index2 = index2 + 1;
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}
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index = index + 1;
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}
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// if v1 is done, let's now deal with v2 left overs
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if(index1>=size1){
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while(index2<size2){
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v[index] = v2[index2];
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index++;
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index2++;
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}
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}else{
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// else means v2 is done. let's now deal with v1 left overs
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while(index1<size1){
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v[index] = v1[index1];
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index++;
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index1++;
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}
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}
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return v;
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}
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vector<int> sortArray(vector<int>& nums) {
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int size = nums.size();
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// base case
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if(size==1){
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return nums;
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}
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// general case
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// split evenly or not evenly does not matter that much in this case
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int mid = size/2;
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vector<int> nums1(mid, 0);
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vector<int> nums2(size-mid, 0);
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for(int i=0;i<mid;i++){
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nums1[i]=nums[i];
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}
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for(int i=mid;i<size;i++){
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// caution: here the index in nums2 has to be i-mid, rather than i; and if we use i here, it will cause a heap buffer overflow issue.
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nums2[i-mid]=nums[i];
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}
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// sort the left half
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nums1 = sortArray(nums1);
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// sort the right half
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nums2 = sortArray(nums2);
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return mergeArrays(nums1, nums2);
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}
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};
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