每日一练-leetcode
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。
来源:力扣(LeetCode)
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
此题和我上次发的三数之和是一样 思路就不写了,代码入下:
class Solution { public List<List<Integer>> fourSum(int[] nums, int target) { List<List<Integer>> quadruplets = new ArrayList<List<Integer>>(); 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 (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) { break; } if (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; } if (nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) { break; } if (nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) { continue; } int left = j + 1, right = 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--; } } } } return quadruplets; } }