LeetCode 1. Two Sum

最新推荐文章于 2020-10-21 10:33:51 发布

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最新推荐文章于 2020-10-21 10:33:51 发布

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分类专栏： Multiple Passes LeetCode 题解

本文链接：https://blog.youkuaiyun.com/github_34333284/article/details/51102967

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本文探讨了在整数数组中寻找两个数相加等于特定目标值的问题，提供了两种高效解决方案：双指针法和哈希映射法，并讨论了二叉搜索树中的变化问题。

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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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

First Method: two pointers.

vector<int> twoSum(vector<int>& nums, int target) {
    sort(nums.begin(), nums.end());  // suppose the input array is not sorted.
    if(nums.size() < 2) return {};
    int forward = 0;
    int backward = nums.size() - 1;
    while(forward < backward) {
        int sum = nums[forward] + nums[backward];
        if(sum == target) {
            return {forward, backward};
        } else if(sum > target) {
            backward--;
        } else {
            forward++;
        }
    }
}  // in this case, time complexity is O(nlgn)... space is O(1)

Second Method: use hashMap

// if the input is not sorted.
vector<int> twoSumII(vector<int>& nums, int target) {
    unordered_map<int, int> dataToIndex;
    for(int i = 0; i < nums.size(); ++i) {
        auto iter = dataToIndex.find(target - nums[i]);
        if(iter == dataToIndex.end()) {
            dataToIndex.insert(make_pair(nums[i], i));
        } else {
            return {i, iter->second};
        }
    }
}  // time complexity O(n), space complexity  O(n).

Variation:

Input is a binary search tree, find two sum.

bool twoSum(TreeNode* root, int target) {
  int val_1 = 0, val_2 = 0;
  stack<TreeNode*> leftHalf;
  stack<TreeNode*> rightHalf;
  bool leftSide = false, rightSide = false;
  while(1) {
    TreeNode* leftNode = root;
    while(leftSide == false) {
      if(leftNode) {
        leftHalf.push(leftNode);
        leftNode = leftNode->left;
      } else {
        if(leftHalf.empty()) leftSide = true;
        else {
          TreeNode* tmp = leftNode.top();
          leftNode.pop();
          val_1 = tmp->val;
          leftNode = leftNode->right;
          leftSide = true;
        }
      }
    }
  TreeNode* rightNode = root;
  while(rightSide == false) {
    if(rightNode) {
        rightHalf.push(rightNode);
        rightNode = rightNode->right;
    } else {
      if(rightHalf.empty()) rightSide = true;
      else {
        TreeNode* tmp = rightNode.top();
        rightNode.pop();
        val_2 = tmp->val;
        rightNode = tmp->left;
        rightSide = true;
      }
    }
  }
  if(val_1 != val_2 && target = val_1 + val_2) return true;
  else if(val_1 + val_2 < target) leftSide = false;
  else rightSide = false;
  if(val_1 >= val_2) return false;
  }
}