18. 四数之和
题目描述
给定一个包含 n 个整数的数组 nums 和一个目标值 target,判断 nums 中是否存在四个元素 a,b,c 和 d ,使得 a + b + c + d 的值与 target 相等?找出所有满足条件且不重复的四元组。
注意:答案中不可以包含重复的四元组。
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 = [], target = 0 输出:[]
提示:
0 <= nums.length <= 200-109 <= nums[i] <= 109-109 <= target <= 109
解法
“排序 + 双指针”实现。
Python3
class Solution: def fourSum(self, nums: List[int], target: int) -> List[List[int]]: res = [] if nums is None or len(nums) < 4: return res n = len(nums) nums.sort() for i in range(n - 3): if i > 0 and nums[i] == nums[i - 1]: continue for j in range(i + 1, n - 2): if j > i + 1 and nums[j] == nums[j - 1]: continue p, q = j + 1, n - 1 while p < q: if p > j + 1 and nums[p] == nums[p - 1]: p += 1 continue if q < n - 1 and nums[q] == nums[q + 1]: q -= 1 continue t = nums[i] + nums[j] + nums[p] + nums[q] if t == target: res.append([nums[i], nums[j], nums[p], nums[q]]) p += 1 q -= 1 elif t < target: p += 1 else: q -= 1 return res
Java
class Solution { public List<List<Integer>> fourSum(int[] nums, int target) { int n; if (nums == null || (n = (nums.length)) < 4) { return Collections.emptyList(); } Arrays.sort(nums); List<List<Integer>> res = new ArrayList<>(); for (int i = 0; i < n - 3; ++i) { if (i > 0 && nums[i] == nums[i - 1]) { continue; } for (int j = i + 1; j < n - 2; ++j) { if (j > i + 1 && nums[j] == nums[j - 1]) { continue; } int p = j + 1, q = n - 1; while (p < q) { if (p > j + 1 && nums[p] == nums[p - 1]) { ++p; continue; } if (q < n - 1 && nums[q] == nums[q + 1]) { --q; continue; } int t = nums[i] + nums[j] + nums[p] + nums[q]; if (t == target) { res.add(Arrays.asList(nums[i], nums[j], nums[p], nums[q])); ++p; --q; } else if (t < target) { ++p; } else { --q; } } } } return res; } }
JavaScript
/** * @param {number[]} nums * @param {number} target * @return {number[][]} */ var fourSum = function (nums, target) { let len = nums.length; let res = []; if (len < 4) return []; nums.sort((a, b) => a - b); for (i = 0; i < len - 3; i++) { if (i > 0 && nums[i] === nums[i - 1]) continue; if (nums[i] + nums[len - 1] + nums[len - 2] + nums[len - 3] < target) continue; if (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) break; for (j = i + 1; j < len - 2; j++) { if (j > i + 1 && nums[j] === nums[j - 1]) continue; let left = j + 1, right = len - 1; while (left < right) { if (nums[i] + nums[j] + nums[left] + nums[right] === target) { res.push([nums[i], nums[j], nums[left], nums[right]]); while (nums[left] === nums[left + 1]) left++; left++; while (nums[right] === nums[right - 1]) right--; right--; continue; } else if (nums[i] + nums[j] + nums[left] + nums[right] > target) { right--; continue; } else { left++; continue; } } } } return res; };