【Leet code 1】 Two sum

最新推荐文章于 2023-03-27 16:12:47 发布

Zlase

最新推荐文章于 2023-03-27 16:12:47 发布

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分类专栏： leetcode

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本文解析了在无序和有序数组中寻找两数之和等于特定目标值的算法实现。对于无序数组，采用双层循环暴力求解；对于有序数组，则利用双指针技巧，实现更高效的查找。文章提供了C++代码示例。

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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, and you may not use the same element twice.

Example:

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

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

大致的意思是：

给定一个数组（vector），在传入一个target，找出其中两个数字加和为target的数字的下标，并返还回来。

解析：

这个题是Easy范围的，因为输入的数字都是无序的，所以使用暴力法就可以了，两个for循环遍历搞定。

#include<iostream>
#include<vector>
using namespace std;

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

};




int main() 
{
	//2 7 11 15
	vector<int> nums,out;
	nums.push_back(3);
	nums.push_back(2);
	nums.push_back(4);
//	nums.push_back(15);
	int target = 6;
	Solution so;
	out=so.twoSum(nums, target);
	cout << out[0] << ' ' << out[1] << endl;


	system("pause");
	return 0;
}

拓展一下，如果现在给的是有序的递增数组，该怎么操作呢？

一个指针在前边一个指针在后边，开始叠加，如果两个数字的和大于target，后边的指针向前走，如果两个数字的和小于target，前边的指针向后移动一位。时间复杂度是O(n)。

#include<iostream>
#include<vector>
using namespace std;


class Solution {
public:
	vector<int> twoSum(vector<int>& nums, int target) {
		int p2 = nums.size()-1;
		int p1 = 0;
		vector<int> vec;
		while (p1 != p2)
		{
			if (nums[p1] + nums[p2] == target)
			{
				vec.push_back(p1);
				vec.push_back(p2);
				return vec;
			}
			else if (nums[p1] + nums[p2] > target)
			{
				p2--;
			}
			else {
				p1++;
			}
		}
		return vec;
	}
};



int main() 
{
	//2 7 11 15
	vector<int> nums,out;
	nums.push_back(2);
	nums.push_back(3);
	nums.push_back(11);
	nums.push_back(15);
	int target = 9;
	Solution so;
	out=so.twoSum(nums, target);
	cout << out[0] << ' ' << out[1] << endl;


	system("pause");
	return 0;
}

===================================================

public boolean[] isSecure(int[][] incompatibility, List<Integer>[] containers){
        boolean[] returnNum = new boolean[containers.length];
        for(int i=0;i<containers.length;i++){
            returnNum[i] = true;
        }

        int numIncompatibility = incompatibility.length;
        Map<Integer,List<Integer>> unComfortableMap = new HashMap<>();
        for(int i=0;i<numIncompatibility;i++){
            List<Integer> newList = unComfortableMap.get(incompatibility[i][0]);
            newList.add(incompatibility[i][1]);
            unComfortableMap.put(incompatibility[i][0],newList);
        }

        for(int i=0;i<containers.length;i++){
            List<Integer> listTemp = containers[i];
            for(Integer temp :listTemp){
                if(unComfortableMap.keySet().contains(temp)){
                    for(Integer getNum: unComfortableMap.get(temp) ){
                        if(listTemp.contains(getNum)){
                            returnNum[i] = false;
                            break;
                        }
                    }
                }
            }
        }

        return returnNum;
    }