跟1 差不多，就是多了个一旦相等，就挪边缘的条件， 参考http://blog.csdn.net/linhuanmars/article/details/20588511
public class Solution { public boolean search(int[] A, int target) { if(A==null || A.length==0) return false; int l = 0; int r = A.length-1; while(l<=r) { int m = (l+r)/2; if(target == A[m]) return true; if(A[m]<A[r]) { if(target>A[m] && target<=A[r]) l = m+1; else r = m-1; } else if(A[m]>A[r]) { if(target>=A[l] && target<A[m]) r = m-1; else l = m+1; } else r--; } return false; } }
Search in Rotated Sorted Array II
原文：http://www.cnblogs.com/jiajiaxingxing/p/4434108.html