标签：before res 重复元素 二分法 int turn element targe NPU
Suppose an array sorted in ascending order 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.
Example 1:
Input: [3,4,5,1,2] Output: 1
Example 2:
Input: [4,5,6,7,0,1,2] Output: 0
class Solution { public int findMin(int[] nums) { int res = nums[0]; for(int i = 0; i < nums.length-1; i++){ if(nums[i+1] < nums[i]) return nums[i+1]; } return res; } }
第一个右<左的时候就是要找的。
class Solution { public int findMin(int[] nums) { // int res = nums[0]; // for(int i = 0; i < nums.length-1; i++){ // if(nums[i+1] < nums[i]) return nums[i+1]; // } // return res; if(nums.length == 1) return nums[0]; int left = 0; int right = nums.length - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] < nums[right]) { right = mid; } else { left = mid+1; } } return nums[left]; } }
也可以用二分法，和查找target相反，
A[mid] < A[right]，则区间[mid,right]一定递增，断层一定在左边A[mid] > A[right]，则区间[left,mid]一定递增，断层一定在右边nums[mid] == nums[right]，这种情况不可能发生，因为数组是严格单调递增的，不存在重复元素153. Find Minimum in Rotated Sorted Array
标签：before res 重复元素 二分法 int turn element targe NPU
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11516045.html
