04 Sorting

题目 1：MaxProductOfThree

A non-empty zero-indexed array A consisting of N integers is given. The product of triplet (P, Q, R) equates to A[P] * A[Q] * A[R] (0 ≤ P < Q < R < N).

For example, array A such that:

A[0] = -3 A[1] = 1 A[2] = 2 A[3] = -2 A[4] = 5 A[5] = 6

contains the following example triplets:

(0, 1, 2), product is −3 * 1 * 2 = −6

(1, 2, 4), product is 1 * 2 * 5 = 10

(2, 4, 5), product is 2 * 5 * 6 = 60

Your goal is to find the maximal product of any triplet.

Write a function:

int solution(vector<int> &A);

that, given a non-empty zero-indexed array A, returns the value of the maximal product of any triplet.

Assume that:

N is an integer within the range [3..100,000];

each element of array A is an integer within the range [−1,000..1,000].

Complexity:

expected worst-case time complexity is O(N*log(N));

expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

#include <algorithm>// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;int solution(vector<int> &A) {// write your code in C++11int n=A.size();sort(A.begin(),A.end());int sum1=A[0]*A[1]*A[n-1];int sum2=A[n-1]*A[n-2]*A[n-3];   return sum1>sum2?sum1:sum2;
}

题目 2：Distinct

Write a function

int solution(vector<int> &A);

that, given a zero-indexed array A consisting of N integers, returns the number of distinct values in array A.

Assume that:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [−1,000,000..1,000,000].

For example, given array A consisting of six elements such that:

A[0] = 2 A[1] = 1 A[2] = 1A[3] = 2 A[4] = 3 A[5] = 1

the function should return 3, because there are 3 distinct values appearing in array A, namely 1, 2 and 3.

Complexity:

expected worst-case time complexity is O(N*log(N));

expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

#include <algorithm>
#include <set>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;int solution(vector<int> &A) {// write your code in C++11//sortint n=A.size();sort(A.begin(),A.end());int count=0;for(int i=0;i<n;i++){if(i==0||A[i]!=A[i-1])count++;}return count;//set/* #define FOR(i,n) for(int i=0;i<(int)n;++i)int n=A.size();set<int> st;FOR(i,n){st.insert(A[i]);}return st.size();*/
}

题目 3：Triangle

A zero-indexed array A consisting of N integers is given. A triplet (P, Q, R) is triangular if 0 ≤ P < Q < R < N and:

A[P] + A[Q] > A[R],

A[Q] + A[R] > A[P],

A[R] + A[P] > A[Q].

For example, consider array A such that:

A[0] = 10 A[1] = 2 A[2] = 5 A[3] = 1 A[4] = 8 A[5] = 20

Triplet (0, 2, 4) is triangular.

Write a function:

int solution(vector<int> &A);

that, given a zero-indexed array A consisting of N integers, returns 1 if there exists a triangular triplet for this array and returns 0 otherwise.

Assume that:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

Complexity:

expected worst-case time complexity is O(N*log(N));

expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

// you can use includes, for example:#include <algorithm>// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;int solution(vector<int> &A) {// write your code in C++11int n=A.size();sort(A.begin(),A.end());for(int i=0;i<n-2;i++){if(A[i+2]-A[i+1]<A[i])return 1;}return 0;
}

题目 4：NumberOfDiscIntersections

We draw N discs on a plane. The discs are numbered from 0 to N − 1. A zero-indexed array A of N non-negative integers, specifying the radiuses of the discs, is given. The J-th disc is drawn with its center at (J, 0) and radius A[J].

We say that the J-th disc and K-th disc intersect if J ≠ K and the J-th and K-th discs have at least one common point (assuming that the discs contain their borders).

The figure below shows discs drawn for N = 6 and A as follows:

A[0] = 1 A[1] = 5 A[2] = 2 A[3] = 1 A[4] = 4 A[5] = 0

There are eleven (unordered) pairs of discs that intersect, namely:

discs 1 and 4 intersect, and both intersect with all the other discs;

disc 2 also intersects with discs 0 and 3.

Write a function:

int solution(vector<int> &A);

that, given an array A describing N discs as explained above, returns the number of (unordered) pairs of intersecting discs. The function should return −1 if the number of intersecting pairs exceeds 10,000,000.

Assume that:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [0..2,147,483,647].

Complexity:

expected worst-case time complexity is O(N*log(N));

expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

int solution(vector<int> &A) {// write your code in C++11int n=A.size();vector<pair<long,int> > plane;for(int i=0;i<n;i++){plane.push_back(make_pair((long)i-A[i],-1));plane.push_back(make_pair((long)i+A[i],1));}sort(plane.begin(),plane.end());long count=0,allinsert=0;for(int i=0;i<2*n;i++){if(plane[i].second==-1){allinsert+=count;if(allinsert>1e7)return -1;count++;}elsecount--;}return allinsert;
}

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