leetcode——TwoSum、3Sum

TwoSum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

解析：尋找序列中唯一的和爲目標數的兩個元素，並返回下標。做法一是直接暴力搜索，但是時間效率低下。另一種可行的做法是先將元素排序，然後用兩個指針指向頭尾，通過改變指針去尋找目標元素。由於需要返回下標，因此可以先用pair存儲元素與其初始下標。具體實現代碼如下，時間複雜度爲O(nlogn)：

class Solution {
public:
	static bool cmp(pair<int, int>& a, pair<int, int>& b)
	{
		return a.first < b.first;
	}
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> v;
        vector<pair<int, int> >	num;
        int n = nums.size(), j = 0, k = n - 1;
        for(int i=0; i<n; i++)
        	num.push_back(make_pair(nums[i], i));
        sort(num.begin(), num.end(), cmp);
        while(j < k)
        {
        	if(num[j].first + num[k].first < target)	j ++;
        	else if(num[j].first + num[k].first > target)	k --;
        	else
        	{
        		v.push_back(num[j].second);
        		v.push_back(num[k].second);
        		break;
        	}
		}
		return v;
    }
};

還有一種實現比較簡易，利用map容器存儲元素，之後只需要對每個元素nums[i]判斷是否存元素target-nums[i]即可，代碼如下：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> ans;
        map<int, int> M;
        int n = nums.size();
        for(int i=0; i<n; i++)
        	M[nums[i]] = i+1;
        for(int i=0; i<n; i++)
        	if(M[target - nums[i]] != i+1 && M[target - nums[i]] > 0)
        	{
        		ans.push_back(i);
        		ans.push_back(M[target - nums[i]]-1);
        		break;
			}
		return ans;
    }
};

3Sum

Given an array S of n integers, are there elements a, b, c in S 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.

For example, given array S = [-1, 0, 1, 2, -1, -4],

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

解析：根據TwoSum可以很容易得到3Sum的解法，排序之後，對每個元素nums[i]，在序列nums[i+1]到nums[n-1]中尋找和爲0-nums[i]的兩個元素。具體代碼如下，時間複雜度爲O(n^2)：

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