［Index］Search in Rotated Sorted Array series

最新推荐文章于 2020-11-25 20:39:50 发布

原创最新推荐文章于 2020-11-25 20:39:50 发布 · 357 阅读

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CC 4.0 BY-SA版权

本文介绍了一种在旋转数组中寻找特定目标值的搜索算法。该算法通过二分查找的方式实现，即使数组被旋转过也能有效地找到目标元素的位置，如果未找到则返回-1。文章详细解释了算法的实现思路，并提供了一个Java示例代码。

Good Luck !

//(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

//You are given a target value to search. If found in the array return its index, otherwise return -1.
//Duplicate Allowed!
public class Solution {
    public int search(int[] nums, int target) {
        int l = 0;
        int r = nums.length - 1;
        while(l <= r){
            int m = (l+r)/2;
            if(nums[m] == target)
                return m;
            if(nums[m] > nums[r]){
                if(nums[l] <= target && target < nums[m]){
                    r = m - 1;
                }else{
                    l = m + 1;
                }
                }else if(nums[m] < nums[r]){
                    if(nums[m] < target  && target <= nums[r]){
                        l = m + 1;
                    }else{
                        r = m - 1;
                    }
                }else{
                    r -- ;
                }
        }
        return -1;
    }
}

//Find the minimum element.
//Duplicate Allowed
public class Solution {
    public int findMin(int[] nums) {
        public int findMin(int[] nums) {
        if(nums[0] < nums[nums.length-1])
            return nums[0];
        int r = nums.length-1;
        int l = 0;
        int MIN = nums[0];
        while(l <= r){
            int m = (l+r)/2;
            MIN = Math.min(MIN,nums[m]);
            if(nums[m] > nums[r]){
                l = m + 1;
                
            }else if(nums[m] < nums[r]){
                r = m - 1;
            }else{
                r -- ;
            }
        }
        return MIN;
    }
    }
}