3Sum & 3Sum Closest & 4Sum

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简单延续2sum的做法,用循环,不过代码比较长。原本很多地方写了break,后来发现都不应该加。

class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        vector<vector<int> > res;        int n=num.size();        if (n<=2) return res;        sort(num.begin(),num.end());        int pret=-999;        for (int i=0; i<n-2; i++) {            if (num[i]==num[i-1] && i>0) {                continue;            }            int target = 0-num[i];            map<int, int> m;            for (int j=i+1; j<n; j++) {                while (num[j]==num[j-1] && j<n && pret==target) {                    j++;                }                if (j==n) break;                if (m.find(num[j])!=m.end()) {                    vector<int> cur;                    cur.push_back(num[i]);                    cur.push_back(m[num[j]]);                    cur.push_back(num[j]);                    res.push_back(cur);                    pret=target;                }else {                    m[target-num[j]]=num[j];                }            }        }        return res;    }};

3Sum Closest

O(n^2),内层循环和Container With Most Water的思路相同 

class Solution {public:    int threeSumClosest(vector<int> &num, int target) {        int n=num.size();        if (!n) return 0;        sort(num.begin(),num.end());        int min=9999;        int res=0;        for (int i=0; i<n; i++) {            int j=i+1, k=n-1;            while (j<k) {                int cur=num[i]+num[j]+num[k];                if (abs(cur-target)<min) {                    min=abs(cur-target);                    res=cur;                }                if (cur==target) {                    return target;                }else if(cur>target) {                    k--;                }else {                    j++;                }            }        }        return res;    }};

4Sum

O(n^3)

use sort erase and unique to remove duplicates in a vector

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        int n=num.size();        vector<vector<int>> res;        if (!n) return res;        sort(num.begin(), num.end());        for (int i=0; i<n-3; i++) {            for (int j=i+1; j<n-2; j++) {                int k=j+1, l=n-1;                while (k<l) {                    int cur=num[i]+num[j]+num[k]+num[l];                    if (cur==target) {                        vector<int> temp;                        temp.push_back(num[i]);                        temp.push_back(num[j]);                        temp.push_back(num[k]);                        temp.push_back(num[l]);                        res.push_back(temp);                        do                            k++;                        while (num[k]==num[k-1]);                        do                            l--;                        while (num[l]==num[l+1]);                    }else if (cur>target) {                        do                            l--;                        while (num[l]==num[l+1]);                    }else {                        do                            k++;                        while (num[k]==num[k-1]);                    }                }            }         }        sort(res.begin(), res.end() );        res.erase( unique( res.begin(), res.end() ), res.end() );        return res;    }};




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