一:Find Minimum in Rotated Sorted Array

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).

Find the minimum element.

You may assume no duplicate exists in the array.

数组中没有重复元素。

代码:

  1. class Solution {
  2. public:
  3. int findMin(vector<int>& nums) {
  4.  
  5. int low = ;
  6. int high = nums.size()-;
  7.  
  8. while(low<=high){
  9.  
  10. int mid = low + (high-low)/;
  11. if(nums[mid] == nums[high]){
  12. high--;
  13. }else if(nums[mid] > nums[high]){
  14. low = mid+;
  15. }else{
  16. high = mid;
  17. }
  18. }
  19.  
  20. return nums[low];
  21. }
  22. };

二:Find Minimum in Rotated Sorted ArrayII

与一不同的是,其允许有重复元素

  1. class Solution {
  2. public:
  3. int findMin(vector<int>& nums) {
  4.  
  5. int low = ;
  6. int high = nums.size()-;
  7.  
  8. while(low<=high){
  9.  
  10. int mid = low + (high-low)/;
  11. if(nums[mid] == nums[high]){
  12. high--;
  13. }else if(nums[mid] > nums[high]){
  14. low = mid+;
  15. }else{
  16. high = mid;
  17. }
  18. }
  19.  
  20. return nums[low];
  21. }
  22. };

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