Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e.,0 1 2 4 5 6 7might become4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 二、分析 这题难度有,因为他默认的是数组中所有的元素是不同的。只有分为两种情况...
}privateintsearch(int[] nums,intlow,inthigh,inttarget) {if(low >high)return-1;intmid = (low + high) / 2;if(nums[mid] ==target)returnmid;if(nums[mid] <nums[high]) { //后半部分有序if(nums[mid] < target && target <=nums[high]) //在有序部分,则继续二分此部分returnsearch(num...
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 search. If found in the array return its index, otherwise return -1. You may assume no duplicate exists in the ...
Sample Solution: Java Code: // Define the Main class.publicclassMain{// Define a method to check if there is a pair of elements in the array// that sum up to 'x'.staticbooleansum_pair(intarr_int[],intn,intx){intk;// Find the pivot point where the array is rotated.for(k=0;...
Find Minimum in Rotated Sorted Array II 理论上这题应该也用二分法,碰到起点终点相同的情况就只能进行遍历,但是那样代码会更复杂并且效率上并不见得提高多少,所以这里我直接采用遍历的方法。 实现代码: javapublic class Solution { public int findMin(int[] num) { ...
第一次二分:找到最小数的位置,参考 find minimum number in rotated sorted array 第二次二分:确定 target 在左侧区间还是右侧,用一个普通的二分法即可找到。 class Solution: """ @param A: an integer rotated sorted array @param target: an integer to be searched @return: an integer """ def sear...
There is an integer arraynumssorted in ascending order (with distinct values). Prior to being passed to your function,numsis rotated at an unknown pivot indexk(0 <= k < nums.length) such that the resulting array is[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ....
本题是Search in Rotated Sorted Array的变形。在Search in Rotated Sorted Array中可以通过nums[start]<=nums[mid]或nums[mid]<=nums[end]判断两边是否为有序。但是在这里出现一个问题:比如左半边,如果nums[start]==nums[mid]它是有序吗?对于: ...
You may assume no duplicate exists in the array. 原问题链接:https://leetcode.com/problems/search-in-rotated-sorted-array/ 问题分析 这个问题在之前的文章里有讨论过。总的来说是基于这么一个思路。当我们将一个排序后的数组循环移位之后,其实它就构成了两个递增的段。我们用二分查找法去查找元素的时候...
in); while(in.hasNextInt()) { int num = in.nextInt(); if (num < 0) break; arrayList.add(new Integer(num)); } int result = processArray(arrayList); System.out.println(result); } } 8 changes: 4 additions & 4 deletions 8 src/arpit/Lecture7BinarySearch/BinarySearch.java Original ...