leetcode 97: Search in Rotated Sorted Array
Search in Rotated Sorted ArrayMar
3 '12
Suppose a sorted array 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
).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
public class Solution { public int search(int[] A, int target) { // Start typing your Java solution below // DO NOT write main() function // 8,9,10, 1,2,3,4,5,6,7 target 9 int low=0, high=A.length-1; int mid = low + (high-low)/2; while(low<=high) { mid = low + (high-low)/2; if(A[mid]==target) return mid; else if(A[mid] > target) { if(A[mid] <= A[high]){ // high end is sorted. high = mid-1; } else { if(A[low] > target) { low = mid + 1; } else { high = mid - 1; } } } else { if(A[low]<=A[mid] ) { // low end is sorted low = mid + 1; } else { if( A[high] < target) { high = mid-1; } else { low = mid+1; } } } } return -1; } }
first try wrong result. will miss in certain situation.
public class Solution { public int search(int[] A, int target) { // Start typing your Java solution below // DO NOT write main() function // 8,9,10, 1,2,3,4,5,6,7 target 9 int low=0, high=A.length-1; int mid = low + (high-low)/2; while(low<=high) { mid = low + (high-low)/2; if(A[mid]==target) return mid; else if(A[mid] > target) { if(A[low]<= target) { high = mid-1; } else { low = mid + 1; } } else { if(A[high] >= target) { low = mid + 1; } else { high = mid - 1; } } } return -1; } }