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].解题思路:
使用hashmap, key为元素值, value为元素在数组中的下标. 从头开始遍历，假设遍历的元素为解的右半部分，则我们只需要考虑左半部分的值是否已经出现过，若已出现过，则根据hashmap，返回该下标和当前下标.
解法一:
时间复杂度o(n) 空间复杂度o(n)
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] ans = new int[2];
Map<Integer, Integer> mp = new HashMap<Integer, Integer>();
for(int i = 0; i < nums.length; i++) {
int d = nums[i];
if(mp.containsKey(target - d)) {
ans[1] = i;
ans[0] = mp.get(target - d);
return ans;
} else {
mp.put(d, i);
}
}
return ans;
}
}解法二:
时间复杂度o(n) 空间复杂度o(n)
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> mp;
for(int d : nums) {
mp[d]++;
}
vector<int> ans(2, -1);
int l = INT_MIN, r = INT_MIN;
for(int d : nums) {
cout<<"d="<<d<<" "<<mp[target - d]<<endl;
mp[d]--;
if(mp.find(target - d) != mp.end() && mp[target - d] > 0) {
l = d;
r = target - d;
break;
}
mp[d]++;
}
if(l == INT_MIN && r == INT_MIN) return ans;
cout<<l<<" "<<r<<endl;
int cnt = 0;
for(int i = 0; i < nums.size() && cnt < 2; i++) {
if(nums[i] == l) {
ans[cnt] = i;
cnt++;
} else if(nums[i] == r) {
ans[cnt] = i;
cnt++;
}
}
return ans;
}
};