1. 二分查找
(1) 有序数组查找插入位置: 主要是终止条件的判断,如果查找不到需要被范围的插入位置为begin
public: int searchInsert(vector<int>& nums, int target) { int len = nums.size(); return binarySearch(nums, target, 0, len-1); } private: int binarySearch(vector<int>& nums, int target, int begin, int end) { if (begin > end) { return begin; } int mid = (begin + end) / 2; if (target == nums[mid]) return mid; else if (target < nums[mid]) { return binarySearch(nums, target, begin, mid - 1);} else { return binarySearch(nums, target, mid+1, end);} }
(2) rotated array 寻找最小值
第一种情况: 不包含重复数字,
class Solution { public: int findMin(vector<int>& nums) { int len = nums.size(); return bsearch(nums, 0, len-1); } private: int bsearch(const vector<int> nums, int s, int e){ if (s >= e) return nums[s]; int mid = (s + e) / 2; // 在左侧数组上 if (nums[mid] > nums[e]) return bsearch(nums, mid+1, e); // 在右侧数组上 else return bsearch(nums, s, mid); } };
(3) rotated array 寻找某个值,允许重复:
分成三种情况,(1) 是递增数组 (2) num[s] == num[mid] = nums[e] 用O(n)的时间顺序查找
(3)两段数组的类型, 二分查找,先判断mid是在哪一段数组上
class Solution { public: bool search(vector<int>& nums, int target) { int len = nums.size(); if(len == 0) return false; return bsearch(nums, target, 0, len - 1); } private: bool bsearch(const vector<int>& nums, int target, int s, int e){ if ( s > e) { return false; } if (cornercase(nums, s, e)){ return asearch(nums, target, s, e); } if (nums[s] >= nums[e]){ int mid = (s + e) / 2; if (nums[mid] == target) return true; if (nums[mid] >= nums[s]) { // mid在左边序列 // target比mid大一定在右边序列 if (target > nums[mid]) return bsearch(nums, target, mid+1, e); // t比mid小,当t大于end则一定在左侧 else if (target > nums[e]) return bsearch(nums, target, s, mid-1); else if (target < nums[e]) return bsearch(nums, target, mid+1, e); else return true; } if (nums[mid] < nums[s]) {// mid在右边序列 if (target < nums[mid]) return bsearch(nums, target, s, mid-1); else if(target > nums[e]) return bsearch(nums, target, s, mid-1); else if(target < nums[e]) return bsearch(nums, target, mid+1, e); else return true; } } return false; } bool cornercase(const vector<int>& nums, int s, int e){ int mid = (s + e) / 2; if (nums[s] == nums[e] && nums[mid] == nums[s]){ return true; } if (nums[s] <= nums[e] && nums[mid] >= nums[s]){ return true; } return false; } bool asearch(const vector<int>& nums, int target, int s, int e){ for (int i = s; i <=e; i++){ if (nums[i] == target) return true; } return false; } };
原文地址:https://www.cnblogs.com/cookcoder-mr/p/11080055.html
时间: 2024-08-05 20:33:02