19.1.26 [LeetCode15]3Sum

Given an array nums of n integers, are there elements a, b, c in nums such that a+ b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

题意

在给定数组中找三个数和为0

题解

用了set储存答案防止重复，很慢

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        set<vector<int>> ans;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0)break;
            int j = i + 1, k = nums.size() - 1;
            while (j < k) {
                if (nums[j] + nums[k] == 0 - nums[i]) {
                    vector<int>tmp;
                    tmp.push_back(nums[i]);
                    tmp.push_back(nums[j]);
                    tmp.push_back(nums[k]);
                    ans.insert(tmp);
                    j++,k--;
                }
                else if (nums[j] + nums[k] < 0 - nums[i])j++;
                else k--;
            }
        }
        vector<vector<int>> _ans;
        while (!ans.empty()) {
            auto tmp = *(ans.begin());
            ans.erase(ans.begin());
            _ans.push_back(tmp);
        }
        return _ans;
    }
};

于是换了一下……

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> ans;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0)break;
            if (i > 0 && nums[i] == nums[i - 1])continue;
            int j = i + 1, k = nums.size() - 1;
            while (j < k) {
                if (nums[j] + nums[k] == 0 - nums[i]) {
                    vector<int>tmp;
                    tmp.push_back(nums[i]);
                    tmp.push_back(nums[j]);
                    tmp.push_back(nums[k]);
                    while (j < k&&nums[j] == nums[j + 1])j++;
                    while (j < k&&nums[k] == nums[k - 1])k--;
                    ans.push_back(tmp);
                    j++, k--;
                }
                else if (nums[j] + nums[k] < 0 - nums[i])j++;
                else k--;
            }
        }
        return ans;
    }
};

差得好多哦-。-