[LintCode] 63 Search in Rotated Sorted Array II 解题报告

Description
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.


Example
Given [1, 1, 0, 1, 1, 1] and target = 0, return true.
Given [1, 1, 1, 1, 1, 1] and target = 0, return false.


思路
因为有duplicates，所以没法用二分法做。因为你在碰到一个数的时候没法知道到底是往左还是往右搜索。只能for循环过一遍。

Code

public class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
    public boolean search(int[] A, int target) {
        // write your code here
        for (int i = 0; i < A.length; i++) {
            if (A[i] == target) {
                return true;
            }
        }
        return false;
    }
}