【LC總結】K Sum (Two Sum I II/3Sum/4Sum/3Sum Closest)

Two Sum

Problem

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are NOT zero-based.

Notice

You may assume that each input would have exactly one solution

Example

numbers=[2, 7, 11, 15], target=9

return [1, 2]

Challenge

Either of the following solutions are acceptable:

O(n) Space, O(nlogn) Time
O(n) Space, O(n) Time

Note

須要返回的是index，不能從新sort，因此，不能用雙指針法，儘可能用HashMap作。

Solution

Brute Force

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] res = {-1, -1};
        if (numbers.length < 2) return res;
        for (int i = 0; i < numbers.length; i++) {
            for (int j = i+1; j < numbers.length; j++) {
                if (numbers[i]+numbers[j] == target) {
                    res[0] = i+1;
                    res[1] = j+1;
                    return res;
                }
            }
        }
        return res;
    }
}

HashMap

public class Solution {
    public int[] twoSum(int[] A, int target) {
        int[] res = {-1, -1};
        if (A == null || A.length < 2) return res;
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < A.length; i++) {
            if (map.containsKey(target-A[i])) {
                res[0] = map.get(target-A[i])+1;
                res[1] = i+1;
                return res;
            }
            else map.put(A[i], i);
        }
        return res;
    }
}

Two Sum II

Problem

Given an array of integers, find how many pairs in the array such that their sum is bigger than a specific target number. Please return the number of pairs.

Example

Given numbers = [2, 7, 11, 15], target = 24. Return 1. (11 + 15 is the only pair)

Challenge

Do it in O(1) extra space and O(nlogn) time.

Note

找符合條件的pair總數。
Key:

1. 雙指針

2. 區間

Solution

public class Solution {
    public int twoSum2(int[] nums, int target) {
        if (nums == null || nums.length == 0) return 0;
        Arrays.sort(nums);
        int count = 0;
        int left = 0, right = nums.length-1;
        while (left < right) {
            if (nums[left]+nums[right] > target) {
                count += right-left;
                right--;
            } else {
                left++;
            }
        }
        return count;
    }
}

3Sum

Problem

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.

Notice

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.

Example

For example, given array S = {-1 0 1 2 -1 -4}, A solution set is:

(-1, 0, 1)
(-1, -1, 2)

Note

考慮邊界，長度<3，爲空，等。
對給定數組排序。
定一個指針i，從頭開始循環推動。若重複，則跳過。
i以後的範圍用雙指針left和right表示。
若三個指針的數字之和爲0，加入結果數組。雙指針繼續移動，若重複，則跳過。
若三個指針的數字之和小於0，左指針後移。
若三個指針的數字之和大於0，右指針前移。

Solution

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] A) {
        ArrayList<ArrayList<Integer>> res = new ArrayList();
        if (A.length < 3) return null;
        Arrays.sort(A);
        for (int i = 0; i <= A.length-3; i++) {
            int left = i+1, right = A.length-1;
            if (i != 0 && A[i] == A[i-1]) continue;
            while (left < right) {
                int sum = A[i]+A[left]+A[right];
                if (sum == 0) {
                    ArrayList<Integer> temp = new ArrayList();
                    temp.add(A[i]);
                    temp.add(A[left]);
                    temp.add(A[right]);
                    res.add(temp);
                    left++;
                    right--;
                    while (left < right && A[left] == A[left-1]) left++;
                    while (left < right && A[right] == A[right+1]) right--;
                }
                else if (sum < 0) left++;
                else right--;
            }
        }
        return res;
    }
}

3Sum Closest

Problem

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.

Notice

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

Example

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).

Challenge

O(n^2) time, O(1) extra space

Note

不要求unique triplets，因此不用判斷duplicate了。
定指針i從數組頭部向後推移，在i的右邊創建左右指針left和right，計算三指針和。
用左右指針夾逼找和的最小值並更新。

Solution

public class Solution {
    public int threeSumClosest(int[] A ,int target) {
        int min = Integer.MAX_VALUE - target;
        if (A == null || A.length < 3) return -1;
        Arrays.sort(A);
        for (int i = 0; i < A.length-2; i++) {
            int left = i+1, right = A.length-1;
            while (left < right) {
                int sum = A[i]+A[left]+A[right];
                if (sum == target) return sum;
                else if (sum < target) left++;
                else right--;
                min = Math.abs(min-target) < Math.abs(sum-target) ? min : sum;
            }
        }
        return min;
    }
}

4Sum

Description
Notes
Testcase
Judge
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.

Notice

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.

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)

Note

同理，頭部兩個指針向後推移，後面創建左右指針夾逼，找到四指針和爲目標值的元素。

Solution

public class Solution {
    public ArrayList<ArrayList<Integer>> fourSum(int[] A, int target) {
        int n = A.length;
        ArrayList<ArrayList<Integer>> res = new ArrayList();
        Arrays.sort(A);
        for (int i = 0; i < n-3; i++) {
            if (i != 0 && A[i] == A[i-1]) continue;
            for (int j = i+1; j <= n-3; j++) {
                if (j != i+1 && A[j] == A[j-1]) continue;
                int left = j+1, right = n-1;
                while (left < right) {
                    int sum = A[i]+A[j]+A[left]+A[right];
                    if (sum == target) {
                        ArrayList<Integer> temp = new ArrayList();
                        temp.add(A[i]);
                        temp.add(A[j]);
                        temp.add(A[left]);
                        temp.add(A[right]);
                        res.add(temp);
                        left++;
                        right--;
                        while (left < right && A[left] == A[left-1]) left++;
                        while (left < right && A[right] == A[right+1]) right--;
                    }
                    else if (sum < target) left++;
                    else right--;
                }
            }
        }
        return res;
    }
}