【LC总结】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

须要返回的是index，不能从新sort，因此，不能用双指针法，尽可能用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

找符合条件的pair总数。
Key:

1. 双指针

2. 区间

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

考虑边界，长度<3，为空，等。
对给定数组排序。
定一个指针i，从头开始循环推动。若重复，则跳过。
i以后的范围用双指针left和right表示。
若三个指针的数字之和为0，加入结果数组。双指针继续移动，若重复，则跳过。
若三个指针的数字之和小于0，左指针后移。
若三个指针的数字之和大于0，右指针前移。

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，因此不用判断duplicate了。
定指针i从数组头部向后推移，在i的右边创建左右指针left和right，计算三指针和。
用左右指针夹逼找和的最小值并更新。

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，头部两个指针向后推移，后面创建左右指针夹逼，找到四指针和为目标值的元素。

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;
    }
}