leetcode--3Sum

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本文介绍了一种解决3Sum问题的有效算法。通过先排序再使用双指针技术，避免了重复解并确保了找到所有唯一三元组。该算法适用于整数数组中寻找三个数相加为零的所有组合。

题目：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]
]

题目分析：重复的元素的考虑：-1 -1 2 符合且按从小到大顺序排列，考虑：返回的list没有重复。

One：用函数：

public class Solution {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    
    public List<List<Integer>> threeSum(int[] nums) {
        int len = nums.length;
        if(nums==null||len<3) return res;
        Arrays.sort(nums);
        for(int i=0;i<len-2;i++){
            if(i>0&&nums[i]==nums[i-1])continue;
            twoSum(nums,i+1,len-1,-nums[i]);
        }
        return res;
    }
    public void twoSum(int[] nums,int begin,int end,int target){
        
        while(begin < end){
            if(nums[begin]+nums[end]==target){
                List<Integer> list = new ArrayList<Integer>();
                list.add(-target);
                list.add(nums[begin]);
                list.add(nums[end]);
                res.add(list);
                while(begin < end && nums[begin] == nums[begin+1]){begin++;}
                while(begin < end && nums[end] == nums[end-1]){end--;}
                begin++;
                end--;
            }else if(nums[begin]+nums[end]<target){
                begin++;
            }else{
                end--;
                
            }
        }
    } 
}

Two：函数写进整个程序

public List<List<Integer>> threeSum(int[] nums) {
    List<List<Integer>> res = new ArrayList<>();
    Arrays.sort(nums);
    for (int i = 0; i + 2 < nums.length; i++) {
        if (i > 0 && nums[i] == nums[i - 1]) {              // skip same result
            continue;
        }
        int j = i + 1, k = nums.length - 1;  
        int target = -nums[i];
        while (j < k) {
            if (nums[j] + nums[k] == target) {
                res.add(Arrays.asList(nums[i], nums[j], nums[k]));
                j++;
                k--;
                while (j < k && nums[j] == nums[j - 1]) j++;  // skip same result
                while (j < k && nums[k] == nums[k + 1]) k--;  // skip same result
            } else if (nums[j] + nums[k] > target) {
                k--;
            } else {
                j++;
            }
        }
    }
    return res;
}

Three：进一步简化

public List<List<Integer>> threeSum(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> list = new ArrayList<List<Integer>>();
    for(int i = 0; i < nums.length-2; i++) {
        if(i > 0 && (nums[i] == nums[i-1])) continue; // avoid duplicates
        for(int j = i+1, k = nums.length-1; j<k;) {
            if(nums[i] + nums[j] + nums[k] == 0) {
                list.add(Arrays.asList(nums[i],nums[j],nums[k]));
                j++;k--;
                while((j < k) && (nums[j] == nums[j-1]))j++;// avoid duplicates
                while((j < k) && (nums[k] == nums[k+1]))k--;// avoid duplicates
            }else if(nums[i] + nums[j] + nums[k] > 0) k--;
            else j++;
        }
    }
    return list;
}