Search in Rotated Sorted Array IIFollow up for “Search in Rotated Sorted Array”:What if duplicates are allowed?Would this affect the run-time complexity? How and why?Write a function to determine if a given target is in the array.
解题思路： 本题基于leetcode 6. 在有序数组旋转后搜索 Search in Rotated Sorted Array，允许有序数组中存在重复。 与上一题（不允许重复）的不同是， nums[m] > nums[l] : (l, m-1)单增 nums[m] <= nums[l] : (m+1, r)不一定单增，因为{1,3,1,1,1} 或{1,1,1,3,1}
此时，，可以将上述条件分开来看 nums[m] < nums[l] : (m+1, r)一定单增 num[m] == nums[l] : 将 l+1，重新递归计算 （当[l, r],将 r-1）
class Solution {public://nums 数组边界为 [l,r)bool searchR(vector<int>& nums,int l, int r, int target) {if (r <= l)return false;int m = (l+r) / 2;if (nums[m] == target)return true;if (nums[l] < nums[m]) {if (target >= nums[l] && target < nums[m])return searchR(nums, l, m, target);elsereturn searchR(nums, m+1, r, target);} else if (nums[l] > nums[m]) {if(target > nums[m] && target <= nums[r-1])return searchR(nums, m+1, r, target);elsereturn searchR(nums, l, m, target);} else {return searchR(nums, ++l, r, target);}}bool search(vector<int>& nums, int target) {return searchR(nums, 0, nums.size(), target);}};
人总是珍惜未得到的，而遗忘了所拥有的
