LintCode : 14. First Position of Target 二分查找，重复数字第一次出现下标

布满荆棘的人生 2021-06-24 16:10 279阅读 0赞

**试题**  
For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Example  
Example 1:  
Input: \[1,4,4,5,7,7,8,9,9,10\]，1  
Output: 0

Explanation: 
    the first index of  1 is 0.

Example 2:  
Input: \[1, 2, 3, 3, 4, 5, 10\]，3  
Output: 2

Explanation: 
    the first index of 3 is 2.

Example 3:  
Input: \[1, 2, 3, 3, 4, 5, 10\]，6  
Output: -1

Explanation: 
    Not exist 6 in array.

Challenge  
If the count of numbers is bigger than 2^32, can your code work properly?  
**代码**  
1、如果相等的话，要查找的值将会在左区间，并且有可能就是mid所在位置。  
2、循环终止条件，对于重复的数最后left和right都会定位到第一个出现的位置，这是如果我们以left==right作为循环终止条件就会陷入死循环。  
3、对于返回值，我们要判断最终查找位置的值是否等于target。

public class Solution {
        /**
         * @param nums: The integer array.
         * @param target: Target to find.
         * @return: The first position of target. Position starts from 0.
         */
        public int binarySearch(int[] nums, int target) {
            int left=0, right=nums.length-1;
            while(left<right){
                int mid = left+(right-left)/2;
                if(nums[mid]==target){
                    right = mid;
                }else if(nums[mid]>target){
                    right = mid-1;
                }else if(nums[mid]<target){
                    left = mid+1;
                }
            }
            if(nums[left]==target)
                return left;
            else
                return -1;
        }
    }