solutions
  • solutions
  • Valid Palindrome
  • String to Integer (atoi)
  • addBinary
  • Longest Palindromic Substring
  • Regular Expression Matching
  • Valid Number
  • Count and say
  • Valid anagram
  • Simplify Path
  • Length of last word
  • Longest Valid Parentheses
  • Valid Parentheses
  • Largest Rectangle in Histogram
  • Evaluate Reverse Polish Notation
    • Morris 算法介绍
  • Binary Tree Preorder Traversal
    • //stack
    • //morris
  • Binary Tree Inorder Traversal
    • //stack
    • //morris
  • Binary Tree Postorder Traversal
    • //stack
  • Binary Tree Level Order Traversal
    • 递归
    • 迭代
  • Binary Tree Zigzag Level Order Traversal
    • 递归
    • 迭代
  • Recover Binary Search Tree
    • Recursive
    • morris
  • Same Tree
  • Symmetric Tree
    • Recursive
    • Iterative
  • Balanced Binary Tree
  • Flatten Binary Tree to Linked List
    • Iterative and recursive
  • Populating Next Right Pointers in Each Node II
    • Iterative
    • Recursive
  • Construct Binary Tree from Preorder and Inorder Traversal
  • Construct Binary Tree from Inorder and Postorder Traversal
  • Unique Binary Search Trees
  • Validate Binary Search Tree
  • Convert Sorted Array to Binary Search Tree
  • Convert Sorted List to Binary Search Tree
  • Minimum Depth of Binary Tree
  • Maximum Depth of Binary Tree
  • Path Sum
  • Path Sum II
  • Binary Tree Maximum Path Sum
  • Populating Next Right Pointers in Each Node
  • Sum Root to Leaf Numbers
  • Merge Sorted Array
  • Merge Two Sorted Lists
    • cpp
    • java
  • Merge k Sorted Lists
    • Priority queue
    • merge sort
  • Insertion Sort List
  • Sort List
  • First Missing Positive
  • Sort Colors
  • Search for a Range
  • Search Insert Position
    • Recursive
    • Iterative
  • Search a 2D Matrix
  • *string 的子集(bit)
  • Subsets
    • Bit
    • Recursive
    • Iterative
  • Subsets II
    • Recursive
    • Iterative
  • Permutations
  • Permutations II
    • Java
  • CPP
  • Combinations
    • Recursive
  • Iterative
  • Letter Combinations of a Phone Number
    • Recursive
  • BFS VS DFS
  • Word Ladder
    • Java
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Search for a Range

class Solution {

public:

vector<int> searchRange(vector<int>& nums, int target) {

vector<int> ret;

int l=binarySearch(nums,target,0,nums.size()-1,true);

int r=binarySearch(nums,target,0,nums.size()-1,false);

ret.push_back(l);

ret.push_back(r);

return ret;

}

int binarySearch(vector<int>& nums, int target,int l,int h, bool left){

if(l>h) return -1;

int m=(h+l)/2;

if(nums[m]==target){

int pos=left ? binarySearch(nums,target,l,m-1,left) : binarySearch(nums,target,m+1,h,left);

return pos==-1 ? m : pos;

}else if(nums[m]>target){

binarySearch(nums,target,l,m-1,left);

}else{

binarySearch(nums,target,m+1,h,left);

}

}

};

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Last updated 5 years ago

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