• 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;
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/nenu/p/15039321.html
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