题目:
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.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
C++
解法1:暴力法
采用二重循环,暴力求解答案
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> ans;
for(int i = 0; i < nums.size(); i++){
for(int j = i + 1; j < nums.size(); j++){
if(nums[i] + nums[j] == target){
ans.clear();
ans.push_back(i);
ans.push_back(j);
return ans;
}
}
}
return ans;
}
};
解法2:map法
新建map储存对应项数值和下标
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> ans;
unordered_map<int, int> m;
for(int i = 0; i < nums.size(); i++){
if(m.find(target - nums[i]) != m.end()){
ans.push_back(m[target - nums[i]]);
ans.push_back(i);
break;
}
m[nums[i]] = i;
}
return ans;
}
};
java
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] ans = new int [2];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < nums.length; i++){
if(map.containsKey(target - nums[i])){
ans[0] = map.get(target - nums[i]);
ans[1] = i;
break;
}
map.put(nums[i], i);
}
return ans;
}
}
Python
class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
map = {}
for i in range(len(nums)):
if target - nums[i] in map:
return [map[target - nums[i]], i]
map[nums[i]] = i;