LeetCode18 四数之和

题目

给你一个由 n 个整数组成的数组 nums ，和一个目标值 target 。请你找出并返回满足下述全部条件且不重复的四元组 [nums[a], nums[
b], nums[c], nums[d]] （若两个四元组元素一一对应，则认为两个四元组重复）： 0 <= a, b, c, d < n ，a、b、c 和 d 互不相同
nums[a] + nums[b] + nums[c] + nums[d] == target

你可以按 任意顺序 返回答案 。

 示例 1： 
输入：nums = [1,0,-1,0,-2,2], target = 0
输出：[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

 示例 2： 
输入：nums = [2,2,2,2,2], target = 8
输出：[[2,2,2,2]]

 提示： 
 1 <= nums.length <= 200 
 -10⁹ <= nums[i] <= 10⁹ 
 -10⁹ <= target <= 10⁹ 

方法

排序法

思路同三数之和，外边再加一层循环

• 时间复杂度：O(n^3)
• 空间复杂度：O(logn)排序的空间复杂度

class Solution {
    private List<List<Integer>> quadruplets = new ArrayList<List<Integer>>();
    public List<List<Integer>> fourSum(int[] nums, int target) {
        if (nums == null || nums.length < 4) {
            return quadruplets;
        }
        Arrays.sort(nums);
        int length = nums.length;
        for (int i = 0; i < length - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            for (int j = i + 1; j < length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                findTwoSum(i,j,nums,target);
            }
        }
        return quadruplets;
    }
    private void findTwoSum(int i,int j,int[] nums,int target){
        int left = j + 1, right = nums.length - 1;
        while (left < right) {
            int sum = nums[i] + nums[j] + nums[left] + nums[right];
            if (sum == target) {
                quadruplets.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                while (left < right && nums[left] == nums[left + 1]) {
                    left++;
                }
                left++;
                while (left < right && nums[right] == nums[right - 1]) {
                    right--;
                }
                right--;
            } else if (sum < target) {
                left++;
            } else {
                right--;
            }
        }
    }
}

排序法优化+剪枝

• 时间复杂度：O()
• 空间复杂度：O()

class Solution {
   class Solution {
    private List<List<Integer>> quadruplets = new ArrayList<List<Integer>>();
    public List<List<Integer>> fourSum(int[] nums, int target) {
        if (nums == null || nums.length < 4) {
            return quadruplets;
        }
        Arrays.sort(nums);
        int length = nums.length;
        for (int i = 0; i < length - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            if ((long) nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
                break;
            }
            if ((long) nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1] < target) {
                continue;
            }
            for (int j = i + 1; j < length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
				//剪枝
                if ((long) nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
                    break;
                }
                if ((long) nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) {
                    continue;
                }
                findTwoSum(i,j,nums,target);
            }
        }
        return quadruplets;
    }
    private void findTwoSum(int i,int j,int[] nums,int target){
        int left = j + 1, right = nums.length - 1;
        while (left < right) {
            int sum = nums[i] + nums[j] + nums[left] + nums[right];
            if (sum == target) {
                quadruplets.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                while (left < right && nums[left] == nums[left + 1]) {
                    left++;
                }
                left++;
                while (left < right && nums[right] == nums[right - 1]) {
                    right--;
                }
                right--;
            } else if (sum < target) {
                left++;
            } else {
                right--;
            }
        }
    }
}