[数据结构与算法]2021-10-11

LeetCode 34

Find First and Last Position of Element in Sorted Array

1.要求:

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.
If target is not found in the array, return [-1, -1].
You must write an algorithm with O(log n) runtime complexity.
example :
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
Input: nums = [], target = 0
Output: [-1,-1]
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array

2.分析：

寻找target在数组里的左右边界，有如下三种处境：
处境一：target 在数组范围的右边或者左边，例如数组{3, 4, 5}，target为2或者数组{3, 4, 5},target为6，返回值为{-1, -1}
处境二：target 在数组范围中，且数组中不存在target，例如数组{3,6,7},target为5，返回值为{-1, -1}
处境三：target 在数组范围中，且数组中存在target，例如数组{3,6,7},target为6，返回值为{1, 1}
考验到要求中时间复杂度为 O(log n)，说明我们应用二分法要找的东西不是target，而是其边界（包括左和右），这里要注意的是，边界问题，分左和右分别进行，不会影响时间复杂度，指针应该从一边走向要找的边界。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int leftBorder = getLeftBorder(nums, target);
        int rightBorder = getRightBorder(nums, target);
        if (leftBorder == -2 || rightBorder == -2) return {-1, -1};
        if (rightBorder - leftBorder > 1) return {leftBorder + 1, rightBorder - 1};
        return {-1, -1};
    }
private:
     int getRightBorder(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        int rightBorder = -2; // 记录一下rightBorder没有被赋值的情况
        while (left <= right) {
            int middle = left + ((right - left) / 2);
            if (nums[middle] > target) {
                right = middle - 1;
            } else { 
                left = middle + 1;
                rightBorder = left;
            }
        }
        return rightBorder;
    }
    int getLeftBorder(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        int leftBorder = -2; // 记录一下leftBorder没有被赋值的情况
        while (left <= right) {
            int middle = left + ((right - left) / 2);
            if (nums[middle] >= target) { 
                right = middle - 1;
                leftBorder = right;
            } else {
                left = middle + 1;
            }
        }
        return leftBorder;
    }
};