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1 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].
//函数参数 为目标数组的应用，和目标和，返回索引数组 vector<int> twoSum(vector<int> &numbers, int target) { //数组元素是键，值是对应的指标 unordered_map<int, int> hash; vector<int> result; for (int i = 0; i < numbers.size(); i++) { //看目标数减去numbers[i] 余数是多少 int numberToFind = target - numbers[i]; //看有没有这个余数的键值 if (hash.find(numberToFind) != hash.end()) { //如果有 返回索引数组 result.push_back(hash[numberToFind]); result.push_back(i); return result; } //不然，更新hasn数组 hash[numbers[i]] = i; } return result; }
167 Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.
Note:
- Your returned answers (both index1 and index2) are not zero-based.
- You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
//也可以使用以上的方法
vector<int> twoSum(vector<int>& numbers, int target) { int l = 0; int r = numbers.size() -1; while(l < r){ if(numbers[l] + numbers[r] == target){ //注意这个初始化方式 vector<int> res{l+1,r+1}; return res; } else if(numbers[l] + numbers[r] > target){ r--; } else{ l++; } } }
560 Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
Example 1:
Input:nums = [1,1,1], k = 2 Output: 2
Note: The length of the array is in range [1, 20,000]. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].
// hashmap + preSum /* 1. Hashmap<sum[0,i - 1], frequency> 2. sum[i, j] = sum[0, j] - sum[0, i - 1] --> sum[0, i - 1] = sum[0, j] - sum[i, j] k sum hashmap-key --> hashmap-key = sum - k 3. now, we have k and sum. As long as we can find a sum[0, i - 1], we then get a valid subsequence which is as long as we have the hashmap-key, we then get a valid subsequence 4. Why don't map.put(sum[0, i - 1], 1) every time ? if all numbers are positive, this is fine if there exists negative number, there could be preSum frequency > 1 */ /* for i, cum is the sum of nums[0] + nums[1] + ... + nums[i]. Assuming there is an item called nums[j] in nums[0] + nums[1] + ... nums[j] + ... + nums[i], if nums[0] + nums[1] + ... + nums[j] equals to cum - k, then, the sum of subarray nums[j + 1] + nums[j + 2] + ... + nums[i] equals to cum - (cum - k) = k, it is one of the subarray which we are looking for. rec[cum - k] is the number of the subarray which begins at index 0 with sum cum - k. */ class Solution { public: int subarraySum(vector<int>& nums, int k) { int cum=0; // cumulated sum map<int,int> rec; // prefix sum recorder int cnt = 0; // number of found subarray rec[0]++; // to take into account those subarrays that begin with index 0 for(int i=0;i<nums.size();i++){ cum += nums[i]; //sum[i, j] = sum[0, j] - sum[0, i - 1] //由于现在的结果是cum 为了凑出k我们还需要 cum - k 的个数 cnt += rec[cum-k]; rec[cum]++; } return cnt; } };
653. Two Sum IV - Input is a BST
Given a Binary Search Tree and a target number, return true if there exist two elements in the BST（二叉搜索树） such that their sum is equal to the given target.
Example 1:
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 9
Output: True
Example 2:
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 28
Output: False
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool findTarget(TreeNode* root, int k) { unordered_set<int> set; return dfs(root, set, k); } bool dfs(TreeNode* root, unordered_set<int>& set, int k){ if(root == NULL)return false; if(set.count(k - root->val))return true; set.insert(root->val); return dfs(root->left, set, k) || dfs(root->right, set, k); } };
更多解法参考：https://leetcode.com/problems/two-sum-iv-input-is-a-bst/discuss/106059/JavaC++-Three-simple-methods-choose-one-you-like