LeetCode : two Sum
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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 same element twice.
我的解决方法:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> result;
int i;
int j;
for(i = 0; i < nums.size(); ++i){
for(j = i+1; j < nums.size(); ++j){
if (nums[i] + nums[j] == target){
result.push_back(i);
result.push_back(j);
return result;
}
}
}
}
太麻烦了!
用hash来解决。就相当于创建python里的一个dictionary,key是array里的每一个数字,value是这个数字在array里的index。然后scan一次array,用target减去这个数字,看得出的差在不在 这个hash的key里面,如果在就返回这个key在hash里对应的value即是在array里的index。
复杂度可以从O(n)变成O(1).
public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { map.put(nums[i], i); } for (int i = 0; i < nums.length; i++) { int complement = target - nums[i]; if (map.containsKey(complement) && map.get(complement) != i) { return new int[] { i, map.get(complement) }; } } throw new IllegalArgumentException("No two sum solution");}
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