Two Sum 两数之和
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the sameelement twice.
Example:
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| Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
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resolution
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int* twoSum(int* nums, int numsSize, int target, int* returnSize){
int i, j;
int *result = NULL;
for(i = 0; i< numsSize -1; i++){
for(j = i + 1; j < numsSize; j++){
if(nums[i] + nums[j] == target){
result = (int *)malloc(sizeof(int)*2);
result[0] = i;
result[1] = j;
*returnSize = 2;
return result;
}
}
}
return result;
}
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class Solution{
public int[] twoSum(int[] nums, int target){
for (int i = 0; i < nums.length; i++ ){
for (int j = i + 1; j < nums.length; j++ ){
if ( nums[i] + nums[j] == target ){
return new int[] {i, j};
}
}
}
throw new IllegalArgumentException("No Solution");
}
}
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Detail:
Time complexity: $O(n^2)$
n x n (row x column )
Space complexity: $O(1)$
Key: C malloc & calloc
Another Good Method:
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| class Solution{
public int[] twoSum(int[] nums, int target){
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++){
int complement = target - num[i];
if(map.containsKey(complement)){
return new [] {map.get(complement), i};
}
map.put(num[i], i);
}
throw new InllegalArgumentException("No Solution");
}
}
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