Find Minimum in rotated Sorted array (smallest number of rotated arrays)

Title Description:

 is 0 1 2 4 5 6 7 4 5 6 7 0 1 2   in the array.

This question "Jian refers to offer" on the original question, directly on the code

Solution:

intFindmin (vector<int>&nums) {    intStart =0; intEnd = Nums.size ()-1;  while(Start <end) {        if(Nums[start] <Nums[end]) Break; intMID = start + (End-start)/2; if(Nums[mid] >=Nums[start]) Start= Mid +1; ElseEnd=mid; }    returnNums[start];}

Reference:https://leetcode.com/discuss/13389/compact-and-clean-c-solution

The above procedure does not take into account the existence of the same element in the array. If considered, the code needs to be modified.

Solution:

intFindmin (vector<int>&nums) {    intStart =0; intEnd = Nums.size ()-1;  while(Start <end) {        if(Nums[start] <Nums[end]) Break; intMID = start + (End-start)/2; if(Nums[mid] >Nums[end]) Start= Mid +1; Else if(Nums[mid] <Nums[end]) End=mid; Else        {            ++start; --end; }    }    returnNums[start];}

Finally, attach a solution to the offer of swords:

intMininorder (vector<int> &nums,intStartintend) {    intMin =Nums[start];  for(inti =1; I < nums.size (); ++i) {if(Nums[i] <min) min=Nums[i]; }    returnmin;}intFindmin (vector<int>&nums) {    intStart =0; intEnd = Nums.size ()-1;  while(Start <end) {        if(Nums[start] <Nums[end]) Break; intMID = start + (End-start)/2; if(Nums[mid] = = Nums[start] && Nums[mid] = =Nums[end]) {            returnMininorder (nums, start, end); }        if(Nums[mid] >=Nums[start]) Start= Mid +1; ElseEnd=mid; }    returnNums[start];}

Ps:

The solution is not necessarily the best in the "Sword Point offer"!!!

Find Minimum in rotated Sorted array (smallest number of rotated arrays)