kSum问题总结

1、2Sum

题目：

方法一：两次迭代

public class TwoSum {

    public static int[] twoSum(int[] nums, int target) {
        int[] indices = {-1,-1};
        for(int i=0; i<nums.length-1; i++ ){
            if(target>=nums[i]){
                for(int k=i+1; k<=nums.length-1; k++){
                    if(nums[k] == (target-nums[i])){
                        indices[0]=i;
                        indices[1]=k;
                        return indices;
                    }
                }

            }
        }
        return indices;

    }
}

方法二：利用HashMap，减少一次迭代

import java.util.HashMap;

public class Two_Sum1 {
    public static int[] twosum(int[] nums, int target){
        int[] result = new int[2];
        if(nums.length < 2) return result;

        HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
        for(int i=0; i<nums.length; i++){
            if(!map.containsKey(target-nums[i])){
                map.put(nums[i],i);
            }else{
                result[0]=map.get(target-nums[i]);
                result[1]=i;
                break;
            }
        }
        return result;
    }
}

2、3Sum

思路分析：数组排序 + twoPointers

public class _3Sum {
    public static List<List<Integer>> threeSum(int[] num){
        Arrays.sort(num);
        List<List<Integer>> res = new LinkedList<>();
        for(int i=0; i<num.length-2; i++){
            if(i==0 || (i>0 && num[i]!= num[i-1])){
                int lo=i+1, hi=num.length-1, sum=0-num[i];
                while(lo<hi){
                    if(num[lo]+num[hi]==sum){
                        res.add(Arrays.asList(num[i],num[lo],num[hi]));
                        while(lo<hi && num[lo] == num[lo+1]) lo++;
                        while(lo<hi && num[hi] == num[hi-1]) hi--;
                        lo++; hi--;
                    }else if(num[lo]+num[hi]<sum) lo++;
                    else hi--;
                }
            }
        }
        return res;
    }
}

3、3Sum Cloest

思路分析：数组排序 + twoPointers

public class _3SumClosest {
    public static int threeSumClosest(int[] nums, int target){
        Arrays.sort(nums);
        int diff = Integer.MAX_VALUE, closest=0;
        for(int i=0;i<nums.length-2;i++){
            int lo=i+1, hi=nums.length-1;
            while(lo<hi){
                int sum = nums[i]+nums[lo]+nums[hi];
                if(sum == target) return target;
                else if(sum > target){
                    if(sum-target<diff){
                        diff = sum - target;
                        closest = sum;
                    }
                    hi--;
                }else{
                    if(target-sum<diff){
                        diff = target - sum;
                        closest = sum;
                    }
                    lo++;
                }
            }
        }
        return closest;
    }
}

4、4Sum

思路分析：数组排序+转化问题为3Sum + 2Sum

public class FourSum {
    public  static List<List<Integer>> fourSum(int[] nums, int target){

        LinkedList<List<Integer>> res = new LinkedList<List<Integer>>();
        if(nums==null || nums.length<4) return res;

        Arrays.sort(nums);
        int len = nums.length;
        int max = nums[len-1];
        if(4*nums[0]>target || 4*max<target) return res;

        for(int i=0; i<len-3; i++){
            int z=nums[i];
            if(i>0 && z==nums[i-1]) continue;
            if(z+3*max<target) continue;
            if(4*z>target) break;
            if(4*z==target) {
                if(i+3<len && nums[i+3]==z) res.add(Arrays.asList(z,z,z,z));
                break;
            }
            threeSum(nums,target-z,i+1,len-1,res,z);
        }
        return res;
    }

    public static void threeSum(int[] nums, int target, int lo, int hi, LinkedList<List<Integer>> fourSumList, int z1){

        if(lo+1>=hi) return;

        int max = nums[hi];
        if(3*nums[lo]>target || 3*max<target) return;

        for(int i=lo; i<hi-1; i++){
            int z=nums[i];
            if(i>lo && z==nums[i-1]) continue;
            if(z+2*max<target) continue;
            if(3*z>target) break;
            if(3*z == target){
                if(i+1<hi && nums[i+2]==z) fourSumList.add(Arrays.asList(z1,z,z,z));
                break;
            }
            twoSum(nums,target-z,i+1,hi,fourSumList,z1,z);
        }
    }

    public static void twoSum(int[] nums, int target, int lo, int hi, LinkedList<List<Integer>> fourSumList, int z1, int z2){
        if(lo>=hi) return;
        if(2*nums[lo]>target || 2*nums[hi]<target) return;

        while(lo<hi){
            int sum = nums[lo]+nums[hi];
            if(sum == target){
                fourSumList.add(Arrays.asList(z1,z2,nums[lo],nums[hi]));

                while(lo<hi && nums[lo]==nums[lo+1]) lo++;
                while(lo<hi && nums[hi]==nums[hi-1]) hi--;
                lo++;hi--;
            }
            if(sum<target) lo++;
            if(sum>target) hi--;
        }
        return;
    }
}

5、kSum

public class KSum {
    public  ArrayList<List<Integer>> kSum(int[] nums, int target, int k, int index){
        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
        if(nums==null || nums.length<k) return res;
        Arrays.sort(nums);
        int len = nums.length;
        int max = nums[len-1];
        if(k*nums[0]>target || k*max<target) {
            return res;
        }
        if(index >= len) {
            return res;
        }
        if(k==2){
            int i=index, j=len-1;
            while(i<j){
                if(nums[i]+nums[j] == target){
                    res.add(Arrays.asList(nums[i],nums[j]));
                    while(i<j && nums[i]==nums[i+1]) i++;
                    while(i<j && nums[j]==nums[j-1]) j--;
                    i++;j--;
                } else if(nums[i]+nums[j]<target) i++;
                else j--;
            } //while循环结束
        }else{
            for (int i = index; i < len - k + 1; i++) {
                ArrayList<List<Integer>> temp = kSum(nums,target-nums[i],k-1,i+1);
                if(temp!=null && temp.size()!=0){
                    for(List<Integer> t : temp){
                        t.add(0,nums[i]);
                    }
                    res.addAll(temp);
                }
                while(i<len-1 && nums[i] == nums[i+1]){
                    i++;
                }
            }//for循环结束
        }
        return res;
    }
}