Problem Description

• Given an array S of n integers, are there any k elements in S such that a[1]+a[2]+..+a[k] = target? Find all unique quadruplets in the array which gives the sum of target.

Some Details

• #1 k=2 and target=0, return the index, the solution is unique
• #15 k=3 and target=0
• #16 k=3, return the sum which is closest to the target
• #18 k=4
• #167 == #1 (O(n) is required)

Solution

#1 
2個數的和。
(1)列舉每一對數，加加看 O(n^2)慢嗎？
(2)如果有序的話，取a[l],a[r]為首尾數，若a[l]+a[r]<target,l++，若a[l]+a[r]>target,r--，否則，找到了。O(n)
正確性是顯然的。那麼排序時間O(nlgn)即為時間複雜度

#15 
3個數的和。
(1)列舉任意三個數，加加看 O(n^3)，太慢了？
(2)a[i]+a[j]=-a[k]，如果a[i]是個常數的話，c+a[j]=-a[k],問題變為2個數的和，先有序O(nlgn)後處理O(n)。怎麼把a[i]變常數？列舉一下i，這樣整體複雜度O(n^2)

#16 
最近的，和#15一樣的題呀。

#18
多列舉一維，也是一個道理。O(n^3)

#167
喵喵喵？#1(2)
 

Code

//隨便貼兩個吧
//#167
class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) 
    {
        vector<int> ans;
        int len=numbers.size();
        int i=0,j=len-1;
        while (i<j)
        {
            if (numbers[i]+numbers[j]>target) j--; else
 

            if (numbers[i]+numbers[j]<target) i++; else
            {
                    ans.push_back(i+1);
                    ans.push_back(j+1);
                    return ans;                
            }
        }
        return ans;
    }
};

//#16
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
       int siz=nums.size(),i=0,j,k;
       int ans=nums[0]+nums[1]+nums[2];
       sort(nums.begin(),nums.end()); 
       while (i<siz-1)
       {
        j=i+1; k=siz-1;
        while (j<k)
        {
            int now=nums[i]+nums[j]+nums[k];
            if (abs(target-now)<abs(target-ans)) ans=now;
            if (now==target) return target;
            if (now>target && j<k) k--; else
            if (now<target && j<k) j++;
            if (j>=k) break;
            //j++; k--;
        }
        i++;
       }
       return ans;        
    }
};