leetcode 7. 在有序可重复数组旋转后搜索 Search in Rotated Sorted Array II
2015-04-27 16:27
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Search in Rotated Sorted Array II
Follow 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); else return 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); else return 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); } };
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