【LeetCode】3Sum
2014-04-01 14:30
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参考:
http://blog.csdn.net/xshalk/article/details/8148422题目描述
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:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
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)
题目分析
题目大意:给定一个数组,存在三个数a,b,c符合a+b+c=0。列出所有可能性。输出要求,abc必须符合a ≤ b ≤ c
思路:
这个可以转换为tow sum问题。逐个遍历数组,取出一个元素a,然后利用twosum问题找出两个相加等于b+c=-a。
对于2sum问题
需要先对数据进行排序。
把数组进行排序,然后首尾各定义一个指针head,tail,
如果target>(num[head]+num[tail]) tail--;
如果target<(num[head]+num[tail]) head++;
如果排序的复杂度为nlogn,那个3sum复杂度为n^2logn
代码示例
/* 编译环境CFree 5.0 博客地址:http://blog.csdn.net/Snowwolf_Yang */ #include #include using namespace std; /* vector > 后一个>前需要加一个空格 */ class Solution { public: vector > threeSum(vector &num) { vector > ret; ret.clear(); sort(num.begin(),num.end()); for(int i=0; i!=num.size();i++){ if(i > 0 && num[i]==num[i-1]) //和前一个相同表示该子问题已经解决 continue; int j,k; j=i+1; k=num.size()-1; while(ji+1&&num[j]==num[j-1]){ //和前一个相同表示该子问题已经解决 j++; continue; } if(k0){ k--; }else if(sum<0){ j++; }else{ vector tmp; tmp.push_back(num[i]); tmp.push_back(num[j]); tmp.push_back(num[k]); ret.push_back(tmp); j++; } } } return ret; } }; void test0() { int arr[] = {0,-1,2,3,-4,5,6,-7,8,9}; vector numbers (arr, arr + sizeof(arr) / sizeof(int) ); Solution so; vector< vector > out = so.threeSum(numbers); int i = 0,j = 0, flag = 1; for(i = 0;i
推荐学习C++的资料
C++标准函数库http://download.csdn.net/detail/chinasnowwolf/7108919
在线C++API查询
http://www.cplusplus.com/
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