codility _ count distinct slices _ min abs sum of two

https://codility.com/programmers/lessons/13

Caterpillar method

Caterpillar还挺形象。。。

题目：CountDistinctSlices Count Distinct Slices

int solution(int M, vector<int> &A) {
    // write your code in C++11
    int right = 0;
    vector<int> buf(M+1, -1);
    int count = 0;
    for(int i=0;i<A.size();i++) {
        while(right<A.size()&&buf[A[right]]==-1) {
            count+=right-i+1;
            if(count>1000000000) return 1000000000;
            buf[A[right]] = right;
            right++;
        }
        if(right==A.size()) break;
        if(right<A.size()&&buf[A[right]]!=-1) {
           int newi = buf[A[right]];
           while(i<newi) {
               buf[A[i]]=-1;
               i++;
           }
           buf[A[right]]=-1;
        }
    }
    return count;    
}

题目：MinAbsSumOfTwo Min abs sum of two

bool cmp(int a,int b)
{
    return abs(a)<abs(b);
}

int solution(vector<int> &A) {
    // write your code in C++11
    sort(A.begin(),A.end(),cmp);
    if(A.size()==0) return 0;
    int res = abs(A[0]*2);
    for(int i=0;i<A.size()-1;i++) {
        if(abs(A[i]+A[i+1])<res)
            res = abs(A[i]+A[i+1]);
        if(abs(A[i]*2)<res)
            res = abs(A[i]*2);
    }
    return min(res,abs(A[A.size()-1]*2));
}