給定一個整數數組,返回兩個數的指數,他們之和等於給定的目標值。
例如:
nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
如下是網上給的時間在O(n),擊敗了97.82%答案:
public class Solution {
public int[] TwoSum(int[] nums, int target) {
int len = nums.Length;
int[] a = new int[2];
Dictionary<int,int> hash = new Dictionary<int,int>();
for (int i=0;i<len;i++) {
int complement = target - nums[i];
if (hash.ContainsKey(complement)) {
a[0]=i;
a[1]=hash[complement];
return a;
}
else {
if (!hash.ContainsKey(nums[i]))
hash.Add(nums[i],i);
}
}
return a;
}
}
但是它只找出了一組值就return了,如果有兩組以上,求不適用。如輸入:{ 1, 2, 4, 5, 6, 7, 0 , -1},目標值6.所以我優化了,給出瞭如下答案:
public class Solution
{
public static int[] TwoSum(int[] nums, int target)
{
int[] data = new int[nums.Length];
int index = 0;
Dictionary<int, int> dic = new Dictionary<int, int>();
for (int i = 0; i < nums.Length; i++)
{
int temp = target - nums[i];
if (dic.ContainsKey(temp))
{
data[index++] = dic[temp];
data[index++] = i;
}
else
{
if (!dic.ContainsKey(nums[i]))
dic.Add(nums[i], i);
}
}
if (index == 0)
{
return null;
}
int[] returnData = new int[index];
for (int i = 0; i < index; i++)
{
returnData[i] = data[i];
}
return returnData;
}
}