leetcode -- 二分查找

题目1.  旋转数组的二分查找

https://leetcode.com/problems/search-in-rotated-sorted-array/

题目简介：一个有序数组，但是经过了旋转， 给一个数字，找到这个数字在数组中位置。

题解：标准二分查找

class Solution {
    public int search(int[] nums, int target) {
        if(nums.length == 0 )return -1;
        int start = 0;
        for(int i=1;i<nums.length;i++){
            if(nums[i] < nums[i-1]){
                start = i;
                break;
            }
        }
        System.out.println(start);
        if(start ==0){
            return partion(nums,0,nums.length-1,target);  //注意，传入的是  nums.length-1
        }else{
            int index = partion(nums,start,nums.length-1,target);
            return  index == -1 ?partion(nums,0,start-1,target):index;
        }
        
    }
    
    int partion(int[] nums,int start,int end,int target){
        int left = start;
        int right = end;
        while(left <= right){  //是小于等于
            int mid = left +(right - left)/2;
            
            if(nums[mid] == target){
                return mid;
            }else if(nums[mid] >target){
                right = mid -1;    // mid -1
            }else {
                left = mid+1;     // mid +1
            }
        }
        return -1;
    }
}

---

题目2 .二分查找，若没有则找合适的插入位置

https://leetcode.com/problems/search-insert-position/

题目简介：二分查找，若没有则找合适的插入位置

题解：略

class Solution {
    public int searchInsert(int[] nums, int target) {
        if(nums.length == 0)return 0;
        return partion(nums,target);
    }
    
    int partion(int[] nums,int target){
        int l = 0,r = nums.length-1;
        int t = 0; 
        while(l <= r){
            int mid = l +(r-l)/2;
            t = mid;                            //mid就是很好的插入位置，因为最后找不到的话 l和r 必然会集中在最靠近的值那里
            if(nums[mid] == target)return mid;
            else if(nums[mid] > target){
                r = mid -1;
            }else{
                l = mid +1;
            }
        }
        return target <nums[t] ? t: t+1 ;
    }
}