[leetcode-81]Search in Rotated Sorted Array II(C)
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问题描述:
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.
分析:
根据这道题的特点,需要找到转折点,其条件是nums[i+1]
bool search(int* nums, int numsSize, int target) { int i; for(i = 0;i<numsSize;i++){ if(nums[i]==target) return true; if(i+1<numsSize && nums[i+1]<nums[i]) break; } int left = i+1; int right = numsSize-1; while(left<=right){ int mid = (left+right)/2; if(nums[mid]==target) return true; else if(nums[mid]<target) left = mid+1; else right = mid-1; } return false;}
代码2:暴力搜索(8ms)
int search(int* nums, int numsSize, int target) { int i; for(i = 0;i<numsSize;i++){ if(nums[i]==target) break; } if(i>=numsSize) return -1; return i;}
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