Leetcode二分查找算法

Python的二分搜索模块

#从左边开始找一个位置bisect.bisect_left(arr, target)#从右边找到一个位置bisect.bisect_right(arr, target)bisect.bisect(arr, target)#有序插入bisect.insort(arr, target)#例：arr:[1，2，2，4，5，6]bisect_left(arr,2) => 1bisect_right(arr, 2) => 3

Search for a Range

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

Given [5, 7, 7, 8, 8, 10] and target value 8,return [3, 4].If the target is not found in the array, return [-1, -1].

Python代码

import bisectclass Solution:    def searchRange(self, nums, target):        re1 = bisect.bisect_left(nums,target)        re2 = bisect.bisect(nums,target)        if re2 == re1:            return [-1,-1]        return [re1, re2-1]

Search in Rotated Sorted Array

Suppose an array sorted in ascending order 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 search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

#先找出旋转点#通过realmid = (mid + rot) % n找出真实的中点class Solution(object):    def search(self, nums, target):        lo, hi = 0, len(nums)-1        while lo < hi:            mid = (lo + hi) // 2            if nums[mid] > nums[hi]:                lo = mid + 1            else:                hi = mid        rot = lo        lo, hi = 0, len(nums)-1        while lo <= hi:            mid = (hi + hi) // 2            realmid = (mid + rot) % len(nums)            if nums[realmid] == target:                return realmid            if nums[realmid] < target:                lo = mid + 1            else:                hi = mid - 1        return -1

public int search(int[] A, int target) {    int lo = 0;    int hi = A.length - 1;    if (hi < 0) {        return -1;    }    while (lo < hi) {        int mid = (lo + hi) / 2;        if (A[mid] == target) return mid;        if (A[lo] <= A[mid]) {            if (target >= A[lo] && target < A[mid]) {                hi = mid - 1;            } else {                lo = mid + 1;            }        } else {            if (target > A[mid] && target <= A[hi]) {                lo = mid + 1;            } else {                hi = mid - 1;            }        }    }    return A[lo] == target ? lo : -1;}