[LeetCode] #18 四數之和
阿新 • • 發佈:2021-10-09
[LeetCode] #18 四數之和
給你一個由 n 個整陣列成的陣列 nums ,和一個目標值 target 。請你找出並返回滿足下述全部條件且不重複的四元組 [nums[a], nums[b], nums[c], nums[d]] :
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 任意順序 返回答案 。
需要再套一層迴圈
class Solution { public List<List<Integer>> fourSum(int[] nums, int target) { List<List<Integer>> list = new ArrayList<>(); int n = nums.length; if(n < 4) return list; Arrays.sort(nums); for(int i = 0; i < n; i++){ if(i > 0 && nums[i] == nums[i - 1]) continue;for(int j = i + 1; j < n; j++){ if(j > i + 1 && nums[j] == nums[j - 1]) continue; int left = j + 1, right = n - 1; while(left < right){ if(nums[i] + nums[j] + nums[left] + nums[right] > target) right--;else if(nums[i] + nums[j] + nums[left] + nums[right] < target) left++; else{ list.add(Arrays.asList(nums[i],nums[j],nums[left],nums[right])); while(left < right && nums[right] == nums[right - 1]) right--; while(left < right && nums[left] == nums[left + 1]) left++; right--; left++; } } } } return list; }
優化
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; } }
知識點:無
總結:無