二分查找

153. Find Minimum in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

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

Find the minimum element.

You may assume no duplicate exists in the array.

Example 1:

Input: [3,4,5,1,2] 
Output: 1

Example 2:

Input: [4,5,6,7,0,1,2]
Output: 0

class Solution {
public:
    int findMin(vector<int>& nums) {
        int LeftIndex = 0;int RightIndex = nums.size() - 1;
        while(LeftIndex < RightIndex){
            int mid = (LeftIndex + RightIndex) / 2;
            if(nums[mid] > nums[RightIndex]){
                LeftIndex = mid +1;
            }
            else{
                RightIndex = mid;
            }
        }
        return nums[RightIndex];
    }
};

二叉树模板：

/**查找第一个与key相等的元素的下标，　如果不存在返回-1　*/
     static int firstEqual(int[] arr,int key){
        int L = 0, R = arr.length - 1; //在[L,R]查找第一个>=key的
        int mid;
        while( L <= R){
            mid = L + (R - L)/2;
            if(arr[mid] >= key)
                R = mid - 1;
            else
                L = mid + 1;
        }
        if(L < arr.length && arr[L] == key)
            return L;
        return -1;
    }

    /**查找第一个大于等于key的元素的下标*/
    static int firstLargeEqual(int[] arr,int key){
        int L = 0, R = arr.length - 1;
        int mid;
        while( L <= R){
            mid = L + (R - L) / 2;
            if(arr[mid] >= key)
                R = mid - 1;
            else
                L = mid + 1;
        }
        return L;
    }


    /**查找第一个大于key的元素的下标 */
    static int firstLarge(int[] arr,int key){
        int L = 0,R = arr.length - 1;
        int mid;
        while(L <= R){
            mid = L + (R - L) / 2;
            if(arr[mid] > key)
                R = mid - 1;
            else
                L = mid + 1;
        }
        return L;
    }


    /**查找最后一个与key相等的元素的下标，　如果没有返回-1*/
     static int lastEqual(int[] arr,int key){
        int L = 0, R = arr.length - 1;
        int mid;
        while( L <= R){
            mid = L + (R - L)/2;
            if(arr[mid] <= key)
                L = mid + 1;
            else
                R = mid - 1;
        }
        if(R >= 0 && arr[R] == key)
            return R;
        return -1;
    }

    /**查找最后一个小于等于key的元素的下标 */
    static int lastSmallEqual(int[] arr,int key){
        int L = 0, R = arr.length - 1;
        int mid;
        while( L <= R){
            mid = L + (R - L) / 2;
            if(arr[mid] <= key)
                L = mid + 1;
            else
                R = mid - 1;
        }
        return R;
    }


    /**查找最后一个小于key的元素的下标*/
    static int lastSmall(int[] arr,int key){
        int L = 0, R = arr.length - 1;
        int mid;
        while(L <= R){
            mid = L + (R - L) / 2;
            if(arr[mid] < key)
                L = mid + 1;
            else
                R = mid - 1;
        }
        return R;
    }
 ———————————————— 
版权声明：本文为优快云博主「zxzxin」的原创文章，遵循CC 4.0 by-sa版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.youkuaiyun.com/zxzxzx0119/article/details/82670761