Leetcode 2SUM-3SUM-4SUM

之前刷过Leetcode 一些题，现在有时间整理一下！

2Sum

题意大概如下：
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].

思路：本题采用双指针，先对给定的数组进行排序，然后一个指针Head在数组开头，另一个Rear指针指向数组最后一个一个元素，如果两个指针所指的元素之和等于目标值，则已找到答案返回，如果大于目标值，则Rear向前指一个;如果小于目标值，则Head向后移动一个元素。注意Head小于Rear的位置。

代码如下：

public class Solution {    static class Node implements Comparable<Node>{        int index;        int val;        Node(int index,int val){            this.index = index;            this.val = val;        }        public int compareTo(Node node){            return this.val - node.val;        }    }    public int[] twoSum(int[] nums, int target) {        int []result = new int[2];        int len  = nums.length;        int head = 0;        int rear = len-1;        int tmp;        Node []nodes = new Node[len];        for(int i=0;i<len;i++){            nodes[i] = new Node(i,nums[i]);        }        Arrays.sort(nodes);        while(head<rear){            tmp = nodes[head].val+nodes[rear].val;            if(tmp==target){                result[0] = nodes[head].index;                result[1] = nodes[rear].index;                return result;            }else if(tmp>target){                rear--;            }else{                head++;            }        }        return result;    }}

运行结果：

能用类似的方法完成3SUM和4SUM，因为在3SUM和4SUM中不需要记录数字的位置，只是返回数值的集合，所以这里还要对2SUM作出一点修改！先把3SUM和4SUM两个问题贴上来！

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]]

3sum 问题就是在数组中找一个数，然后在剩下的数字中完成2SUM的操作;4SUM 问题就是在数组中找一个数，然后在剩余数字中完成3SUM的操作！
下面先把2SUM的代码修改一下！

public void sum2Helper(int[]a,int target,int start,int end,int cur,List<List<Integer>>result){        int head = start;        int rear = end;        int tmp;        while(head<rear) {            tmp = a[head] + a[rear];            if (tmp == target) {                List<Integer> row = new ArrayList<Integer>();                row.add(a[head]);                row.add(a[rear]);                row.add(a[cur]);                result.add(row);                head++;                rear--;                while (head<=rear&&a[head] == a[head - 1]) {                    head++;                }                while (rear>=0&&a[rear] == a[rear + 1]) {                    rear--;                }            } else if (tmp < target) {                head++;            } else {                rear--;            }        }    }

下面给出3SUM完整代码！

public class Solution {    public List<List<Integer>> threeSum(int[] nums) {        Arrays.sort(nums);        return sum3(nums,0);    }    public List<List<Integer>> sum3(int[]a,int target){        int len = a.length;        List<List<Integer>> result = new ArrayList<List<Integer>>();        for(int i=0;i<len-2;i++){            if(i>0&&a[i]==a[i-1]){                continue;            }            sum2Helper(a,target-a[i],i+1,len-1,i,result);        }        return result;    }    public void sum2Helper(int[]a,int target,int start,int end,int cur,List<List<Integer>>result){        int head = start;        int rear = end;        int tmp;        while(head<rear) {            tmp = a[head] + a[rear];            if (tmp == target) {                List<Integer> row = new ArrayList<Integer>();                row.add(a[head]);                row.add(a[rear]);                row.add(a[cur]);                result.add(row);                head++;                rear--;                while (head<=rear&&a[head] == a[head - 1]) {                    head++;                }                while (rear>=0&&a[rear] == a[rear + 1]) {                    rear--;                }            } else if (tmp < target) {                head++;            } else {                rear--;            }        }    }}

在3SUM上进行改进，支持4SUM！

public class Solution {    public List<List<Integer>> fourSum(int[] nums, int target) {        Arrays.sort(nums);        return sum4(nums,target);    }    public List<List<Integer>> sum4(int[]a,int target){        int len = a.length;        List<List<Integer>> result = new ArrayList<List<Integer>>();        for(int i=0;i<len-3;i++){            if(i>0&&a[i]==a[i-1]){                continue;            }            sum3Helper(a,target-a[i],i+1,len-1,i,result);        }        return result;    }    public void sum3Helper(int[]a,int target,int start,int end,int cur,List<List<Integer>>result){        for(int i=start;i<=end;i++){            if(i>start&&a[i-1]==a[i]){                continue;            }            sum2Helper(a,target-a[i],i+1,end,i,cur,result);        }    }    public void sum2Helper(int[]a,int target,int start,int end,int cur3,int cur4,List<List<Integer>>result){        int head = start;        int rear = end;        int tmp;        while(head<rear) {            tmp = a[head] + a[rear];            if (tmp == target) {                List<Integer> row = new ArrayList<Integer>();                row.add(a[head]);                row.add(a[rear]);                row.add(a[cur3]);                if(cur4>=0)                    row.add(a[cur4]);                result.add(row);                head++;                rear--;                while (head<=rear&&a[head] == a[head - 1]) {                    head++;                }                while (rear>=0&&a[rear] == a[rear + 1]) {                    rear--;                }            } else if (tmp < target) {                head++;            } else {                rear--;            }        }    }}

最后给出4SUM 的运行结果！时间是O(n3)