摘要：解題思路題目要求兩個(gè)數(shù)和等于，返回其題目說(shuō)明不會(huì)有重復(fù)情況，所以我們一旦發(fā)現(xiàn)符合情況的，就可以直接結(jié)束循環(huán)并返回。特殊情況就是正好等于，那肯定是最接近的情況，直接返回即可。

Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

1.解題思路

題目要求兩個(gè)數(shù)和等于target，返回其index.題目說(shuō)明不會(huì)有重復(fù)情況，所以我們一旦發(fā)現(xiàn)符合情況的，就可以直接結(jié)束循環(huán)并返回。
利用HashMap,邊遍歷存儲(chǔ)，邊尋找。當(dāng)給定一個(gè)數(shù)n,只要看target-n是否在map中即可,如果存在，則取出其value - 即該數(shù)的下標(biāo)。

2.代碼

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res=new int[2];
        if(nums.length==0) return res;
        HashMap hm=new HashMap();
        for(int i=0;i
3Sum
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: 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]
]

1.解題思路
本題要求三個(gè)數(shù)的和為0，其實(shí)就是固定一個(gè)數(shù)n之后，找兩個(gè)數(shù)和為0-n，所以就轉(zhuǎn)化為求兩個(gè)數(shù)的和，那我們很容易想到使用二分法。
注意點(diǎn)：本題需要排除重復(fù)值，
1）固定的數(shù)下標(biāo)為i,則要從i+1開(kāi)始尋找后兩個(gè)數(shù)；
2）如果數(shù)組中包含重復(fù)的數(shù)，則為了保證結(jié)果不出現(xiàn)重復(fù)，我們每次遇到重復(fù)的數(shù)需要跳過(guò)。
2.代碼
public class Solution {
    List> res =new ArrayList>();
    public List> threeSum(int[] nums) {
      
       Arrays.sort(nums);
      
       for(int i=0;i0&&nums[i]==nums[i-1]) continue; //避免重復(fù)
           twoSum(0-nums[i],nums,i+1,nums.length-1);
       }
       return res;
    }
    private void twoSum(int target,int[] nums, int start,int end){
        
        int i=start,j=end;
        while(i subres=new ArrayList();
            int sum=nums[i]+nums[j];
            if(sum==target){
                subres.add(0-target);
                subres.add(nums[i]);
                subres.add(nums[j]);
                res.add(subres);
                
               do {
                    i++;
                }while(i < end && nums[i] == nums[i-1]);
                do {
                    j--;
                } while(j >= 0 && nums[j] == nums[j+1]);
            }
            else if(sum
4Sum
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: 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]
]

1.解題思路
4Sum,就是在3Sum基礎(chǔ)上再嵌套一層，注意點(diǎn)也是要避免重復(fù)。
2.代碼
public class Solution {
    ArrayList> res = new ArrayList>();
    public List> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        for(int i=0;i0&&nums[i]==nums[i-1]) continue;
              res.addAll(threesum(target-nums[i],nums,i+1));
        }
        return res;
    }
    private List> threesum(int target,int[] nums,int start){
        ArrayList> res = new ArrayList>();
        int first=start-1;
        for(int i=start;istart&&nums[i]==nums[i-1]) continue;
            res.addAll(twosum(target-nums[i],nums,i+1,first));
        }
        return res;
    }
    private List> twosum(int target,int[] nums,int start,int first){
        List> res =new ArrayList>();
        int second=start-1;
        int i=start;
        int j=nums.length-1;
        while(i subres =new ArrayList();
            int sum=nums[i]+nums[j];
            if(sum==target){
                subres.add(nums[first]);
                subres.add(nums[second]);
                subres.add(nums[i]);
                subres.add(nums[j]);
                res.add(subres);
                i++;
                while(i=0&&nums[j]==nums[j+1]){
                    j--;
                }
            }
            else if(sum>target) j--;
            else i++;
            
        }
        return res;
    }
}
3Sum Closest
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).
1.解題思路
與3Sum相似，只不過(guò)本次是要尋找最接近target的3個(gè)數(shù)之和，我們需要維護(hù)minDiff和closetSum兩個(gè)變量。
但是本題題目說(shuō)明不許考慮重復(fù)。 特殊情況就是sum正好等于target，那肯定是最接近的情況，直接返回即可。
2.代碼
public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if(nums.length==0) return 0;
        int minDiff=Integer.MAX_VALUE;
        int closet=0;
        Arrays.sort(nums);
        for(int i=0;itarget){
                    right--;
                }
                else return sum;
            }
        }
        return closet;
    }
}
4Sum II
Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.
Example:

Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

1.解題思路
本題從1個(gè)數(shù)組擴(kuò)展到了4個(gè)數(shù)組，求的是有多少組4數(shù)之和等于target的，我們把問(wèn)題分為兩個(gè)數(shù)一組，假設(shè)當(dāng)前兩個(gè)數(shù)A[i]+B[j],那我們其實(shí)就是要在C和D中求是否存在和為0-A[i]-B[j],如果存在則返回存在的個(gè)數(shù)。想到用HashMap,存儲(chǔ).
2.代碼
public class Solution {
    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
        if(A.length==0||B.length==0||C.length==0||D.length==0) return 0;
        Map map=new HashMap();
        int count=0;
        for(int i=0;i