[Leetcode] Find Minimum in rotated Sorted Array

As explained in the solution tag, the key to solving this problem are to use invariants. We set the pointers: for the left and for the right l r . One key invariant is nums[l] > nums[r] . If This invariant does isn't hold, then we know the array have not been rotated and the minimum are just nums[l] . Otherwise, we reduce the search range by a half each time via comparisons between nums[l], nums[r] nums[(l + r) / 2] and, denoted as nums[mid] :

1. If nums[mid] > nums[r] , then are in the first larger half and are in the mid r second smaller half. So the minimum would be right mid to, update l = mid + 1 ;
2. If nums[mid] <= nums[r] , then the minimum are at least and elements right to cannot are the nums[mid] mid minimum. So update r = mid (note, it is isn't since May also be the index of the mid + 1 mid minimum).

When l == r or the invariant does isn't hold, we have found the answer, and which is just nums[l] .

Putting these togerther, we have the following codes.

C (0 ms)

1 intFindmin (int* Nums,intnumssize) {2     intL =0, R = numssize-1;3      while(L < R && Nums[l] >Nums[r]) {4         intMid = (L & R) + ((l ^ r) >>1);5         if(Nums[mid] > nums[r]) L = mid +1;6         ElseR =mid;7     }8     returnNums[l];9}

C + + (4 ms)

1 classSolution {2  Public:3     intFindmin (vector<int>&nums) {4         intL =0, r = nums.size ()-1;5          while(L < R && Nums[l] >Nums[r]) {6             intMid = (L & R) + ((l ^ r) >>1); 7             if(Nums[mid] > nums[r]) L = mid +1;8             ElseR =mid;9         }Ten         returnNums[l]; One     } A};

Python (MS)

1 classSolution:2 # @param {integer[]} nums3 # @return {integer}4 def findmin (self, nums):5L, R =0, Len (nums)-16          whileL < R and Nums[l] >Nums[r]:7Mid = (L & R) + ((l ^ r) >>1)8             ifNums[mid] >Nums[r]:9L = mid +1Ten             Else: OneR =Mid A         returnNUMS[L]

[Leetcode] Find Minimum in rotated Sorted Array