Search in Rotated Sorted Array II

Leetcode

題意：

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.

解題思路：

updated on 2015.12.25

要考慮大於等於小於太煩雜了，直接一律改成以下

target >= nums[mid] && target <= nums[right]

target >= nums[left] && target <= nums[mid]

public class Solution {
    public boolean search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return false;
        }

        int left = 0;
        int right = nums.length - 1;

        while (left + 1 < right) {
            int mid = left + (right - left) / 2;

            if (nums[mid] < nums[right]) {
                if (target >= nums[mid] && target <= nums[right]) {
                    left = mid;
                } else {
                    right = mid;
                }
            } else if (nums[mid] > nums[right]){
                if (target >= nums[left] && target <= nums[mid]) {
                    right = mid;
                } else {
                    left = mid;
                }
            } else {
                right--;
            }
        }

        if (nums[left] == target || nums[right] == target) {
            return true;
        } else {
            return false;
        }
    }
}

九章模版，差別在於，如果有遇到重複的，則直接start往前移。

public class Solution {
    public boolean search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return false;
        }

        int start = 0;
        int end = nums.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[start] < nums[mid]) {
                if (nums[start] <= target && nums[mid] > target) {
                    end = mid;
                } else {
                    start = mid;
                }
            } else if (nums[start] > nums[mid]){
                if (nums[mid] < target && nums[end] >= target) {
                    start = mid;
                } else {
                    end = mid;
                }
            } else {
                start++;
            }
        }

        if (nums[start] == target) {
            return true;
        } else if (nums[end] == target) {
            return true;
        } else {
            return false;
        }
    }
}