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- Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1]. - code
public int[] TwoSum(int[] nums, int target) {
List<int> indexList = new List<int>();
Hashtable hashTable = new Hashtable();
for (int i = 0; i < nums.Count<int>(); i++)
{
if (hashTable.ContainsKey(target - nums[i]))
{
indexList.Add((int)hashTable[target - nums[i]]);
indexList.Add(i);
break;
}
else if (!hashTable.ContainsKey(nums[i]))
{
hashTable.Add(nums[i], i);
}
}
return indexList.ToArray();
}
- 代码思路:O(n^2)的时间复杂度降为O(n)。第一次遍历是少不了的,第二次遍历查找target-num[i]的O(n)过程用哈希表快速查找某个数值以及相应的下标索引来代替,时间复杂度为O(1)。