【LeetCode】154. Find Minimum in Rotated Sorted Array II (cpp) 题目和测试用例略233 本题作为153题的升级版,难点在于数组中的数可能有重复的。 这种问题很容易想到用二分搜索的思想来解决(毕竟如果暴力求解时间复杂度只有O(N),比O(N)小的就是O(logN)了233)。设l是考虑的左边界,r是右
1)如果nums[mid] < target <= nums[right],说明target在右边区间里,则left = mid + 1; 2)否则在左边区间里,搜索左边区间,right = mid - 1; 3. nums[mid] > nums[right],说明[elft, mid]区间是在左边的递增区间,然后判断target是否在这个左边区间里 1)如果nums[left] <= target < nums[mid],说...
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). 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...
Example 1: Input: nums = [2,5,6,0,0,1,2], target = 0 Output: true Example 2: Input: nums = [2,5,6,0,0,1,2], target = 3 Output: false Noted that: The array may contain dumplicated value, so this is different to Leet Code 33. Search in Rotated Sorted Array In the ...
Search in Rotated Sorted Array I && II Leetcode 对有序数组进行二分查找(下面仅以非递减数组为例): 1. int binarySort(int A[], int lo, int hi, int target) 2. { 3. while(lo <= hi) 4. { 5. int mid = lo + (hi - lo)/2; ...
[leetcode] 81. Search in Rotated Sorted Array II Description Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]). You are given a target value to search. If found in the ...
Can you solve this real interview question? Search in Rotated Sorted Array - There is an integer array nums sorted in ascending order (with distinct values). Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1
Can you solve this real interview question? Search in Rotated Sorted Array - There is an integer array nums sorted in ascending order (with distinct values). Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1
这道是之前那道Search in Rotated Sorted Array 在旋转有序数组中搜索的延伸,现在数组中允许出现重复数字,这个也会影响我们选择哪半边继续搜索,由于之前那道题不存在相同值,我们在比较中间值和最右值时就完全符合之前所说的规律:如果中间的数小于最右边的数,则右半段是有序的,若中间数大于最右边数,则左半段是有...
publicintsearch(int[]nums,inttarget){intlo=0,hi=nums.length-1;while(lo<=hi){intmid=lo+(hi-lo)/2;intnum=nums[mid];//nums [ mid ] 和 target 在同一段if((nums[mid]<nums[0])==(target<nums[0])){num=nums[mid];//nums [ mid ] 和 target 不在同一段,同时还要考虑下变成 -inf 还...