标签:public 重复 忽略 add ++ 判断 return ict nta
public int[] TwoSum(int[] nums, int target) {
Dictionary<int, int> kvs = new Dictionary<int, int>();
for (int i = 0; i < nums.Length; i++)
{
int complement = target - nums[i];
if (kvs.ContainsKey(complement) && kvs[complement] != i)
{
return new int[] { kvs[complement] ,i};
}
//需要对重复值进行判断,若结果包含了重复值,则已经被上面给return了;所以此处对于重复值直接忽略
if (!kvs.ContainsKey(nums[i]))
{
kvs.Add(nums[i], i);
}
}
return new int[] { 0, 0 };
}
}
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标签:public 重复 忽略 add ++ 判断 return ict nta
原文地址:https://www.cnblogs.com/dreamsails/p/13162412.html