33. Search in Rotated Sorted Array旋转数组二分法查询

一句话思路:反正只是寻找一个最小区间,断开也能二分。根据m第一次的落点,来分情况讨论。

一刷报错:

  1. 结构上有根本性错误:应该是while里面包括if,不然会把代码重复写两遍,不好。
 //situation1
        if (nums[mid] > nums[start]) {
            while (start + 1 < mid) {
                mid = start + (end - start) / 2;
                if (nums[mid] == target) {
                    return mid;
                }
                else if (nums[start] <= target && target <= nums[mid]) {
                    end = mid;
                }
                else start = mid;
            }
        }
        
        //situantion2
        else {
            while (start + 1 < mid) {
                mid = start + (end - start) / 2;
                if (nums[mid] == target) {
                    return mid;
                }
                if (nums[mid] <= target && target <= nums[end]) {
                    start = mid;
                }
                else end = mid;
            }
        }

 

  1. int方法的corner case返回值:数组为空,数组长度为0
  2. 居然没有写target== nums[mid]的特殊情况,该死!

一句话总结思路:要把while放在外面,if放在里面。

二刷ac

public class Solution {
    /*
     * @param nums: an integer rotated sorted array
     * @param target: an integer to be searched
     * @return: an integer
     */
    public int search(int[] nums, int target) {
        // write your code here
        if (nums.length == 0 || nums == null) {
            return -1;
        }
        
        int start = 0;
        int end = nums.length - 1;
        int mid;
        
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            //situation1,red line
            if (nums[mid] > nums[start]) {
                if (nums[mid] == target) {
                    end = mid;
                }
                else if (nums[start] <= target && target <= nums[mid]) {
                    end = mid;
                }
                else start = mid;
            }
            //situation2,green line
            else {
                 if (nums[mid] == target) {
                    end = mid;
                }
                if (nums[mid] <= target && target <= nums[end]) {
                    start = mid;
                }
                else end = mid;
            }
        }
        
        if (nums[start] == target) {
            return start;
        }
        if (nums[end] == target) {
            return end;
        }

        return -1;
    }
}
View Code

 

posted @ 2017-11-25 22:07  苗妙苗  阅读(167)  评论(0编辑  收藏  举报