算法：3个数之和15. 3Sum

Posted 2021-11-03 架构师易筋

15. 3Sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Example 2:

Input: nums = []
Output: []

Example 3:

Input: nums = [0]
Output: []

Constraints:

• 0 <= nums.length <= 3000
• -105 <= nums[i] <= 105

两头逼近解法

分析：结果集不能重复，所要要排序，如果现在的数字跟以前相同，则跳过。
这里有三次跳过，可以看下面的 skip the same: 第一个数相同，第二个数相同，第三个数相同；

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        int len = nums.length;
        if (len < 3) return list;
        
        Arrays.sort(nums);
        for (int f = 0; f < len - 2; f++) {
            if (f > 0 && nums[f] == nums[f - 1]) continue; // skip the same
            int lo = f + 1, hi = len - 1;
            int target = 0 - nums[f];
            while (lo < hi) {
                if (nums[lo] + nums[hi] == target) {
                    list.add(Arrays.asList(nums[f], nums[lo], nums[hi]));
                    while (lo < hi && nums[lo] == nums[lo + 1]) lo++; // skip the same
                    while (lo < hi && nums[hi] == nums[hi - 1]) hi--; // skip the same
                    lo++;
                    hi--;
                } else if (nums[lo] + nums[hi] < target) {
                    lo++;
                } else {
                    hi--;
                }
            }
        }
        
        return list;
    }
}