Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
题目大意是：给定一个数组array，和一个目标值target，找出数组中的两个数使得array[i] + array[j] = target,且i < j，返回i和j的下标，下标从1开始。
解题思想：因为array中的数是非排序的，故要对array中的数做一个处理(当然你可以选择排序处理)，这里使用map，建立array中元素和下标的关系(建立hash表, O(1)时间查找)；然后遍历array，在map中查找target - array[i]是否存在；存在则记录其下标(下标要+1，因此数组下标从0开始)。
code：
1 class Solution { 2 public: 3 vector<int> twoSum(vector<int> &numbers, int target) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 map<int, int>h_map; 7 int i; 8 vector<int> indices; 9 //build map 10 for (i = 0; i < numbers.size(); ++i) 11 { 12 h_map.insert(pair<int, int>(numbers[i], i)); 13 } 14 15 int index1 = 0; 16 int index2 = 0; 17 18 for (i = 0; i < numbers.size(); ++i) 19 { 20 //if h_map[target - numbers[i]] exist 21 if (h_map.find(target - numbers[i]) != h_map.end()) 22 { 23 if (i == h_map[target - numbers[i]]) 24 continue; 25 index1 = i; 26 index2 = h_map[target - numbers[i]]; 27 break; 28 } 29 } 30 31 if ( i < numbers.size()) 32 { //pair exist 33 if (index1 > index2) 34 { //swap 35 index1 ^= index2; 36 index2 ^= index1; 37 index1 ^= index2; 38 } 39 indices.push_back(index1 + 1);//non-zero based 40 indices.push_back(index2 + 1); 41 } 42 return indices; 43 } 44 };
test case：
input output
[5,75,25], 100 2, 3
[150,24,79,50,88,345,3], 200 1, 4
[2,1,9,4,4,56,90,3], 8 4, 5
[230,863,916,585,981,404,316,785,88,12,70,435,384,778,887,755,740,337,86,92,325,422,815,650,920,125,277,336,221,847,168,23,677,61,400,136,874,363,394,199,863,997,794,587,124,321,212,957,764,173,314,422,927,783,930,282,306,506,44,926,691,568,68,730,933,737,531,180,414,751,28,546,60,371,493,370,527,387,43,541,13,457,328,227,652,365,430,803,59,858,538,427,583,368,375,173,809,896,370,789], 542 29, 46
[591,955,829,805,312,83,764,841,12,744,104,773,627,306,731,539,349,811,662,341,465,300,491,423,569,405,508,802,500,747,689,506,129,325,918,606,918,370,623,905,321,670,879,607,140,543,997,530,356,446,444,184,787,199,614,685,778,929,819,612,737,344,471,645,726], 789 11, 56
[722,600,905,54,47], 101 4, 5
[210,582,622,337,626,580,994,299,386,274,591,921,733,851,770,300,380,225,223,861,851,525,206,714,985,82,641,270,5,777,899,820,995,397,43,973,191,885,156,9,568,256,659,673,85,26,631,293,151,143,423], 35 40, 46
[286,461,830,216,539,44,989,749,340,51,505,178,50,305,341,292,415,40,239,950,404,965,29,972,536,922,700,501,730,430,630,293,557,542,598,795,28,344,128,461,368,683,903,744,430,648,290,135,437,336,152,698,570,3,827,901,796,682,391,693,161,145], 890 16, 35
[22,391,140,874,75,339,439,638,158,519,570,484,607,538,459,758,608,784,26,792,389,418,682,206,232,432,537,492,232,219,3,517,460,271,946,418,741,31,874,840,700,58,686,952,293,848,55,82,623,850,619,380,359,479,48,863,813,797,463,683,22,285,522,60,472,948,234,971,517,494,218,857,261,115,238,290,158,326,795,978,364,116,730,581,174,405,575,315,101,99], 163 55, 74
[678,227,764,37,956,982,118,212,177,597,519,968,866,121,771,343,561], 295 7, 9
自己写的mian函数
1 int main() 2 { 3 vector<int> numbers; 4 numbers.push_back(2); 5 numbers.push_back(7); 6 numbers.push_back(11); 7 numbers.push_back(15); 8 Solution *ptr = new Solution(); 9 vector<int> indices = ptr->twoSum(numbers, 9); 10 if (indices.size() > 0) 11 { 12 printf("index1 = %d, index2 = %d\n", indices[0], indices[1]); 13 } 14 return 0; 15 }
