LeetCode-Search in Rotated Sorted Array II
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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.
解决方法就是对于A[mid]==A[left]和A[mid]==A[right]单独处理。
public class Solution { public boolean search(int[] nums, int target) { if(nums==null || nums.length==0){ return false; } int left=0; int right=nums.length-1; while(left<=right){ int mid=left+(right-left)/2; if(target<nums[mid]){ if(nums[mid]< nums[right]){// right side is sorted right=mid-1; } else if(nums[mid] == nums[right]){//can‘t tell right is sorted or not, move pointer right--; } else {//left side is sorted if(target<nums[left]){ left=mid+1; } else{ right=mid-1; } } } else if(target>nums[mid]){ if(nums[mid] > nums[left]){//left is sorted left=mid+1; } else if(nums[mid]== nums[left]){// cann‘t tell left side is sorted or not, move pointer left++; } else{//right is sorted if(target>nums[right]){ right=mid-1; } else{ left=mid+1; } } } else{ return true; } } return false; } }
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