《Cracking the Coding Interview》——第17章：普通题——题目12

2014-04-29 00:04

题目：给定一个整数数组，找出所有加起来为指定和的数对。

解法1：可以用哈希表保存数组元素，做到O(n)时间的算法。

代码：

 // 17.12 Given an array of integers and target value, find all pairs in the array that sum up to the target.
 // Use hash to achieve O(n) time complexity. Duplicates pairs are skipped.
 #include <cstdio>
 #include <unordered_map>
 #include <vector>
 using namespace std;

 int main()
 {
     vector<int> v;
     unordered_map<int, int> um;
     int n, i;
     int x, y;
     int target;

     while (scanf("%d", &n) ==  && n > ) {
         scanf("%d", &target);

         v.resize(n);
         for (i = ; i < n; ++i) {
             scanf("%d", &v[i]);
         }

         // duplicate pairs are skipped
         for (i = ; i < n; ++i) {
             um[v[i]] = um[v[i]] + ;
         }

         unordered_map<int, int>::iterator it, it2;
         for (it = um.begin(); it != um.end(); ++it) {
             x = it->first;
             y = target - x;
             if (x > y) {
                 continue;
             }

             --it->second;
             if ((it2 = um.find(y)) != um.end() && it2->second > ) {
                 printf("(%d, %d)\n", x, y);
             }
             ++it->second;
         }

         v.clear();
         um.clear();
     }

     return ;
 }

解法2：先将数组排序，然后用两个iterator向中间靠拢进行扫描。总体时间是O(n * log(n))。

代码：

 // 17.12 Given an array of integers and target value, find all pairs in the array that sum up to the target.
 // O(n * log(n) + n) solution.
 #include <algorithm>
 #include <cstdio>
 #include <vector>
 using namespace std;

 int main()
 {
     vector<int> v;
     int n, i;
     int ll, rr;
     int target;

     while (scanf("%d", &n) ==  && n > ) {
         scanf("%d", &target);

         v.resize(n);
         for (i = ; i < n; ++i) {
             scanf("%d", &v[i]);
         }
         sort(v.begin(), v.end());
         ll = ;
         rr = n - ;

         int sum;
         while (ll < rr) {
             sum = v[ll] + v[rr];
             if (sum < target) {
                 while (ll +  < rr && v[ll] == v[ll + ]) {
                     ++ll;
                 }
                 ++ll;
             } else if (sum > target) {
                 while (rr -  > ll && v[rr] == v[rr - ]) {
                     --rr;
                 }
                 --rr;
             } else {
                 printf("(%d, %d)\n", v[ll], v[rr]);
                 while (ll +  < rr && v[ll] == v[ll + ]) {
                     ++ll;
                 }
                 ++ll;
             }
         }

         v.clear();
     }

     return ;
 }