题解——Leetcode 18. 4Sum 难度:Medium
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Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
本题需要找到数组中四个数满足其和等于目标值,输出所有满足条件的四数组合。
首先对数组按升序排序,共有四层循环,最外层是第一个数,从开头遍历,次外层是第二个数,从第一个数后一位遍历,第三个数从第二个数后一位遍历,第四个数从末尾开始遍历,但第三个数和第四个数只有满足特定条件才会发生位置变化。当第三个数在第四个数左边时,判断四数之和是否等于目标值,如果小于目标值,则第三个数右移一位,如果大于目标值,则第四个数左移一位,如果正好相等,则将四个数加入结果集中。由于经过排序后可能出现相邻数重复的情况,于是加入判断,如果相邻两数相等,则数右移、左移一位直到和左边的数不等。
while (front < back && nums[front] == nums[front - 1]) ++front;
while (front < back && nums[back] == nums[back + 1]) --back;
while(j + 1 < nums.size() && nums[j + 1] == nums[j]) ++j;
while (i + 1 < nums.size() && nums[i + 1] == nums[i]) ++i;
当遍历结束后,输出结果集。
class Solution {public: vector<vector<int>> fourSum(vector<int>& nums, int target) { vector<vector<int> > res; if (nums.empty()) return res; sort(nums.begin(),nums.end()); for (int i = 0; i < nums.size(); i++) { int target_3 = target - nums[i]; for (int j = i + 1; j < nums.size(); j++) { int target_2 = target_3 - nums[j]; int front = j + 1, back = nums.size() - 1; while(front < back) { int two_sum = nums[front] + nums[back]; if (two_sum < target_2) front++; else if (two_sum > target_2) back--; else { res.push_back({nums[i], nums[j], nums[front], nums[back]}); front++, back--; while (front < back && nums[front] == nums[front - 1]) ++front; while (front < back && nums[back] == nums[back + 1]) --back; } } while(j + 1 < nums.size() && nums[j + 1] == nums[j]) ++j; } while (i + 1 < nums.size() && nums[i + 1] == nums[i]) ++i; } return res; }};
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