【LeetCode】解题34：Find First and Last Position of Element in Sorted Array

Problem 34: Find First and Last Position of Element in Sorted Array [Medium]

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 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]

来源：LeetCode

解题思路

• 基本思路是利用二分查找，分别寻找左边界和右边界。
• leftPos()寻找左边界，返回第一个不小于target的数的下标。如果所有数都大于等于target，返回值为0；如果所有数都小于target，返回值为N。显然，当返回N时，数组中不存在target，直接返回结果[-1,-1]。
• rightPos()寻找右边界，返回最后一个不大于target的数的下标。如果所有数都小于等于target，返回值为N-1；如果所有数都大于target，返回值为-1。显然，当leftPos()返回值大于rightPos()返回值时，数组中不存在target，直接返回结果[-1,-1]。
• 正常情况下返回结果[left,right]，其中left和right分别为leftPos()和rightPos()的返回值。

减少执行用时的设计：

• 数组个数为0时直接返回[-1,-1]。
• 在第二阶段寻找右边界时，不用寻找整个数组，可以在leftPos()返回值的基础上，寻找left ~ N-1之间最后一个不大于target的下标，如果left ~ N-1之间的数都大于target，返回值为left-1。

二分查找时间复杂度O(log n)，两次二分查找2*O(log n)，总复杂度依旧为O(log n)。

要点：二分查找、条件判断

Solution (Java)

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int N = nums.length;
        if(N == 0) return new int[] {-1,-1};

        int left = leftPos(nums, 0, N-1, target);
        if(left == N) return new int[] {-1, -1};

        int right = rightPos(nums, left, N-1, target);
        // System.out.println(left+" "+right);
        if(left > right) return new int[] {-1, -1};
        
        return new int[] {left, right};
    }
    // the first index of integers no smaller than target
    private int leftPos(int[] nums, int left, int right, int target){
        if(left > right){
            return left;
        }
        int p = (left + right + 1)/2;
        if(nums[p] >= target){
            return leftPos(nums, left, p-1, target);
        }
        else{
            return leftPos(nums, p+1, right, target);
        }
    }
    // the last index of integers no bigger than target
    private int rightPos(int[] nums, int left, int right, int target){
        if(left > right){
            return right;
        }
        int p = (left + right + 1)/2;
        if(nums[p] <= target){
            return rightPos(nums, p+1, right, target);
        }
        else{
            return rightPos(nums, left, p-1, target);
        }
    }
}