1. Two Sum

题目链接:1. Two Sum

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].

解法一

暴力法,时间复杂度n2, 空间复杂度1

遍历数组,对于每个数num1, 再遍历一遍去寻找num2 = target - num1存不存在。

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

解法二

时间复杂度为nlogn 空间复杂度为nlogn

用set来做,对于每个遍历一遍数组,把每个数num1的解num2 = target - num1存在set里。
再次遍历数组,如果当下的值num1在set里找的到,说明是num1和target-num1存在在数组里,把他们位置找出来就行。

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

题目已经告诉了,可以断定只有一组精确的解,

and you may not use the same element twice.

因为相同的元素不能用两次,所以必须要判重,即是两个数的位置是否相同。

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {

        vector<int> a(0);
        int len = nums.size();
        set<int> s;
        for(int i = 0;i < len;++i){
            s.insert(target - nums[i]);
        }
        for(int i = 0;i < len;++i){
            if(s.find(nums[i])!=s.end()){
                int num1 = nums[i];
                int num2 = target - nums[i];
                int i ;
                int pos = 0;
                for(i = 0;i<len;++i){
                    if(num1 == nums[i]){
                        pos =i;
                        break;
                    }
                }
                a.push_back(pos);
                for(i=0;i<len;++i){
                    if(num2 == nums[i]){
                        pos = i;
                    }
                }
                a.push_back(pos);
                if(a[0] == a[1]){
                    a.clear();
                    continue;
                }
                break;
            }
        }
        return a;
    }
};
Last modification:December 21st, 2019 at 07:58 pm
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