《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) == 1 && n > 0) {
        scanf("%d", &target);
        
        v.resize(n);
        for (i = 0; i < n; ++i) {
            scanf("%d", &v[i]);
        }
        
        // duplicate pairs are skipped
        for (i = 0; i < n; ++i) {
            um[v[i]] = um[v[i]] + 1;
        }
        
        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 > 0) {
                printf("(%d, %d)\n", x, y);
            }
            ++it->second;
        }
        
        v.clear();
        um.clear();
    }
    
    return 0;
}

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) == 1 && n > 0) {
        scanf("%d", &target);
        
        v.resize(n);
        for (i = 0; i < n; ++i) {
            scanf("%d", &v[i]);
        }
        sort(v.begin(), v.end());
        ll = 0;
        rr = n - 1;
        
        int sum;
        while (ll < rr) {
            sum = v[ll] + v[rr];
            if (sum < target) {
                while (ll + 1 < rr && v[ll] == v[ll + 1]) {
                    ++ll;
                }
                ++ll;
            } else if (sum > target) {
                while (rr - 1 > ll && v[rr] == v[rr - 1]) {
                    --rr;
                }
                --rr;
            } else {
                printf("(%d, %d)\n", v[ll], v[rr]);
                while (ll + 1 < rr && v[ll] == v[ll + 1]) {
                    ++ll;
                }
                ++ll;
            }
        }
        
        v.clear();
    }
    
    return 0;
}