Leetcode Hacking Practice -- Solution in C++ Part 1 (發表於10年前)

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第一部分（1/5）

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If you can finish most of these questions you're ready for a new level ...

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If you want to be successfully in technique interview, try solve the problem by yourself.

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Problem Name: 3Sum

Description:

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0?

Find all unique triplets in the array which gives the sum of zero.

Note:

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a <= b <= c)

The solution set must not contain duplicate triplets.

For example, given array S = {-1 0 1 2 -1 -4},

A solution set is:

(-1, 0, 1)

(-1, -1, 2)

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Now comes the solution (in C++) ...

Simplify '3sum' to '2sum' O(n^2). en.wikipedia.org/wiki/3SUM

代碼:


class Solution {

public:

    vector<vector<int> > threeSum(vector<int> &num) {

        vector<vector<int>> res;

        sort(num.begin(), num.end());

        int N = num.size();

        for (int i = 0; i < N-2 && num[i] <= 0; ++i)

        {

            if (i > 0 && num[i] == num[i-1])

                continue; // avoid duplicates

            int twosum = 0 - num[i];

            int l = i + 1, r = N - 1;

            while (l < r)

            {

                int sum = num[l] + num[r];

                if (sum < twosum)

                    l++;

                else if (sum > twosum)

                    r--;

                else {

                    vector<int> triplet;

                    triplet.push_back(num[i]);

                    triplet.push_back(num[l]);

                    triplet.push_back(num[r]);

                    res.push_back(triplet);

                    l++; r--;

                    while (l < r && num[l] == num[l-1]) l++;  // avoid duplicates

                    while (l < r && num[r] == num[r+1]) r--;  // avoid duplicates

                }

            }

        }

        return res;

    }

};

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Problem Name: 3Sum Closest

Description:

Given an array S of n integers, find three integers in S such that the sum is closest to

a given number, target. Return the sum of the three integers.

You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

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Now comes the solution (in C++) ...

Similar to 3Sum, taking O(n^2) time complexity.

代碼:


class Solution {

public:

    int threeSumClosest(vector<int> &num, int target) {

        int res = INT_MAX;

        int N = num.size();

        sort(num.begin(), num.end());

        for (int i = 0; i < N-2; ++i)

        {

            int l = i + 1, r = N - 1;

            while (l < r)

            {

                int threesum = num[i] + num[l] + num[r];

                if (threesum == target) return target;

                else if (threesum < target) l++;

                else r--;

                if (res == INT_MAX || abs(threesum - target) < abs(res - target))

                    res = threesum;

            }

        }

        return res;

    }

};

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Problem Name: 4Sum

Description:

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0?

Find all unique triplets in the array which gives the sum of zero.

Note:

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target?

Find all unique quadruplets in the array which gives the sum of target.

Note:

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a <= b <= c <= d)

The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:

(-1, 0, 0, 1)

(-2, -1, 1, 2)

(-2, 0, 0, 2)

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Now comes the solution (in C++) ...

Similar to 3Sum, 2Sum.

代碼:


class Solution {

public:

    vector<vector<int> > fourSum(vector<int> &num, int target) {

        int N = num.size();

        vector<vector<int> > res;

        if (N < 4) return res;

        sort(num.begin(), num.end());

        for (int i = 0; i < N; ++i)

        {

            if (i > 0 && num[i] == num[i-1]) continue; // avoid duplicates

            for (int j = i+1; j < N; ++j)

            {

                if (j > i+1 && num[j] == num[j-1]) continue; // avoid duplicates

                int twosum = target - num[i] - num[j];

                int l = j + 1, r = N - 1;

                while (l < r)

                {

                    int sum = num[l] + num[r];

                    if (sum == twosum) {

                        vector<int> quadruplet(4);

                        quadruplet[0] = num[i];

                        quadruplet[1] = num[j];

                        quadruplet[2] = num[l];

                        quadruplet[3] = num[r];

                        res.push_back(quadruplet);

                        while (l < r && num[l+1] == num[l]) l++; // avoid duplicates

                        while (l < r && num[r-1] == num[r]) r--; // avoid duplicates

                        l++; r--;

                    }

                    else if (sum < twosum) l++;

                    else r--;

                }

            }

        }

        return res;

    }

};

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Problem Name: Add Binary

Description:

Given two binary strings, return their sum (also a binary string).

For example,

a = "11"

b = "1"

Return "100".