leetcode做题总结，题目Search in Rotated Sorted Array I/II 2012/03/02－12/04/09

从类似45678123这样的有序数组中寻找对应的target，思路是还是按照普通的二分查找，然后再求出中间值mid后将其转换对应到数组中的index，然后进行比较或者返回。

public int search(int[] A, int target) {
        if(A.length==0)return -1;
        if(A.length==1){
            if(A[0]==target)return 0;
            else return -1;
        }
        int qian=1;
        for(int i=0;i<A.length-1;i++){
            if(A[i]>A[i+1])break;
            qian=i+2;
        }
        int hou=A.length-qian;
        int i=0,j=A.length-1,chan;
        while(i<=j){
            int mid=(i+j)/2;
            if(mid<hou)chan=mid+qian;
            else chan = mid-hou;
            //if(hou==0)chan=mid;
            if(A[chan]==target)return chan;
            if(A[chan]>target) j=mid-1;
            else i=mid+1;
        }
        return -1;
    }

第二题在第一题的基础上添加了数组中可能有重复元素，问是否影响复杂度，我觉得会降低复杂度，因为对于相同长度的数组，如果有重复数，相应的二分的命中几率就会大，次数就会少。

public boolean search(int[] A, int target) {
        if(A.length==0)return false;
        if(A.length==1){
            if(A[0]==target)return true;
            else return false;
        }
        int qian=1;
        for(int i=0;i<A.length-1;i++){
            if(A[i]>A[i+1])break;
            qian=i+2;
        }
        int hou=A.length-qian;
        int i=0,j=A.length-1,chan;
        while(i<=j){
            int mid=(i+j)/2;
            if(mid<hou)chan=mid+qian;
            else chan = mid-hou;
            //if(hou==0)chan=mid;
            if(A[chan]==target)return true;
            if(A[chan]>target) j=mid-1;
            else i=mid+1;
        }
        return false;
    }

Update 2015/08/23:下面是新的代码，注意数组判断长度是否为0和判断对象是否为空一样重要。

public class Solution {
    /** 
     *@param A : an integer rotated sorted array
     *@param target :  an integer to be searched
     *return : an integer
     */
    public int search(int[] A, int target) {
        // write your code here
        if(A.length==0)return -1;  
        
        int i=0,j=A.length-1;
        int s = A[i];
        int e = A[j];
        while(i<=j){ 
            int mid=(i+j)/2;  
            if (A[mid] == target)
                return mid;
            else if (A[mid] >= A[i]){
                if (target >= A[i] && target < A[mid])
                    j = mid - 1;
                else
                    i = mid + 1;
            } else {
                if (target <= A[j] && target > A[mid])
                    i = mid + 1;
                else
                    j = mid - 1;
            }
        }  
        return -1; 
    }
}

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.length==0)return false;  
        
        int i=0,j=A.length-1;
        int s = A[i];
        int e = A[j];
        while(i<=j){ 
            int mid=(i+j)/2;  
            if (A[mid] == target)
                return true;
            else if (A[mid] > A[i]){
                if (A[i] <= target && target < A[mid])
                    j = mid - 1;
                else
                    i = mid + 1;
            } else if (A[mid] < A[i]){
                if (target <= A[j] && A[mid] < target)
                    i = mid + 1;
                else
                    j = mid - 1;
            } else
                i++;
        }  
        return false;
    }
}