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Search in Rotated Sorted Array II

any int param repeat loop return i++ turn through

O(N)

if element can repeat, the worst case, you cannot throw away any section. eg. [1, 1, 1, 1, 0, 1, 1] target = 0, you cannot throw any section. We have to loop through to check.

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 if (A == null || A.length == 0) { return false; } for (int i = 0; i < A.length; i++) { if (A[i] == target) { return true; } }
return false; } }

Search in Rotated Sorted Array II