LintCode 1536: Find First and Last Position of Element in Sorted Array

本文介绍了一种在已排序整数数组中查找目标值起始和结束位置的方法，通过二分查找算法实现，确保了算法的时间复杂度为O(logn)。如果目标值不存在于数组中，则返回[-1,-1]。

Find First and Last Position of Element in Sorted Array

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example
Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

解法1：Binary Search
代码同步更新在
https://github.com/luqian2017/Algorithm

class Solution {
public:
    /**
     * @param nums: the array of integers
     * @param target: 
     * @return: the starting and ending position
     */
    vector<int> searchRange(vector<int> &nums, int target) {
        int n = nums.size();
        if (n == 0) return {-1, -1};
        
        int start = 0, end = n - 1;
        int firstPos = 0, endPos = 0;
        //find first position of target
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] >= target) {
                end = mid;
            } else { // if (nums[mid] < target) {
                start = mid;
            } 
        }
        
        if (nums[start] == target) firstPos = start;
        else if (nums[end] == target) firstPos = end;
        else return {-1, -1};
        
        start = 0, end = n - 1;
        //find last position of target
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] <= target) {
                start = mid;
            } else {
                end = mid;
            } 
        }

        if (nums[end] == target) endPos = end;
        else if (nums[start] == target) endPos = start;
        else return {-1, -1};
        
        return {firstPos, endPos};
    }
};