[leetcode]二分查找总结】的更多相关文章

自从做完leetcode上的三道关于二分查找的题后,我觉得它是比链表找环还恶心的题,首先能写出bugfree代码的人就不多,而且可以有各种变形,适合面试的时候不断挑战面试者,一个程序猿写代码解决问题的能力都能在这个过程中考察出来. 在有序数组中寻找等于target的数的下标,没有的情况返回应该插入的下标位置 :http://oj.leetcode.com/problems/search-insert-position/ public int searchInsert(int[] A, int t…
https://oj.leetcode.com/problems/search-for-a-range/就是一个二分查找,没事练练手 public class Solution { public int[] searchRange(int[] A, int target) { int a[]=new int[2]; int ans=bSearch(A,target); if(ans==-1) { a[0]=-1; a[1]=-1; return a; } else { int beg=ans;…
Search for a Range 1.最简单的想法,用最普通的二分查找,找到target,然后向左右扩张,大量的重复的target,就会出现O(n)效率. class Solution { public int[] searchRange(int[] A, int target) { ]; int a= bserch(A,target); ) {ans[]=-;ans[]=-; return ans;} //left ; &&A[b]==target) b--; ans[]=b+; b…
Search in Rotated Sorted Array II Total Accepted: 18500 Total Submissions: 59945My Submissions Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a functi…
本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie Search Insert Position Total Accepted: 14279 Total Submissions: 41575 Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be i…
Search in Rotated Sorted Array Total Accepted: 28132 Total Submissions: 98526My Submissions Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to s…
模板 #2: int binarySearch(vector<int>& nums, int target){ if(nums.size() == 0) return -1; int left = 0, right = nums.size(); while(left < right){ // Prevent (left + right) overflow int mid = left + (right - left) / 2; if(nums[mid] == target){ r…
模板 #1: int binarySearch(vector<int>& nums, int target){ if(nums.size() == 0) return -1; int left = 0, right = nums.size() - 1; while(left <= right){ // Prevent (left + right) overflow int mid = left + (right - left) / 2; if(nums[mid] == targe…
模板 #3: int binarySearch(vector<int>& nums, int target){ if (nums.size() == 0) return -1; int left = 0, right = nums.size() - 1; while (left + 1 < right){ // Prevent (left + right) overflow int mid = left + (right - left) / 2; if (nums[mid] ==…
目录 33/81搜索旋转排序数组 34在排序数组中查找元素的第一个和最后一个位置 35搜索插入位置 74搜索二维矩阵 300最长上升子序列,354俄罗斯套娃信封问题 33/81搜索旋转排序数组 假设按照升序排序的数组在预先未知的某个点上进行了旋转. ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] ). 思路: 设置左右指针开始进入二分循环 判断中间的数是否为target,否则通过比较nums[mid] < nums[right]确定是右边有序还是左边有序…