【剑指offer】排序篇-含题目代码思路解析

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param numbers int整型vector 
     * @return int整型
     */
    int duplicate(vector<int>& numbers) {
    vector<int> newarray(numbers.size());
    if(numbers.size()==0){
        return -1;
    }
    for(int i=0;i<numbers.size();i++){
        newarray[numbers[i]]+=1;
        printf("%d",newarray[numbers[i]]);
        if(newarray[numbers[i]]==2){
            return numbers[i];
          
        }
    }
    return -1;
    }
}; 

class Solution {
public:
    int mod=1000000007;
    int mergeSort(int left,int right,vector<int>& data, vector<int>& temp){
        //停止划分,即划分至长度为1时
        if(left>=right){return 0;}
        //mid
        int mid=(right+left)/2;
        int count=mergeSort(left, mid, data, temp)+mergeSort(mid+1, right, data, temp);
        count%=mod;
        
        int i=left;int j=mid+1;
        for(int k=left;k<=right;k++){
            temp[k]=data[k];
        }
        for(int k=left;k<=right;k++){
            
        }

    }
    int InversePairs(vector<int> data) {
        int n=data.size();
        vector<int> temp(n);
        mergeSort(0,n-1, data,temp);
    }
};