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中等
数组
二分查找

34. 在排序数组中查找元素的第一个和最后一个位置

English Version

题目描述

给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target，返回 [-1, -1]。

你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。

 

示例 1：

输入：nums = [5,7,7,8,8,10], target = 8
输出：[3,4]

示例 2：

输入：nums = [5,7,7,8,8,10], target = 6
输出：[-1,-1]

示例 3：

输入：nums = [], target = 0
输出：[-1,-1]

 

提示：

  • 0 <= nums.length <= 105
  • -109 <= nums[i] <= 109
  • nums 是一个非递减数组
  • -109 <= target <= 109

解法

方法一：二分查找

我们可以进行两次二分查找，分别查找出左边界和右边界。

时间复杂度 $O(\log n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。

Python3

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        l = bisect_left(nums, target)
        r = bisect_left(nums, target + 1)
        return [-1, -1] if l == r else [l, r - 1]

Java

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int l = search(nums, target);
        int r = search(nums, target + 1);
        return l == r ? new int[] {-1, -1} : new int[] {l, r - 1};
    }

    private int search(int[] nums, int x) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int l = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
        int r = lower_bound(nums.begin(), nums.end(), target + 1) - nums.begin();
        if (l == r) {
            return {-1, -1};
        }
        return {l, r - 1};
    }
};

Go

func searchRange(nums []int, target int) []int {
	l := sort.SearchInts(nums, target)
	r := sort.SearchInts(nums, target+1)
	if l == r {
		return []int{-1, -1}
	}
	return []int{l, r - 1}
}

TypeScript

function searchRange(nums: number[], target: number): number[] {
    const search = (x: number): number => {
        let [left, right] = [0, nums.length];
        while (left < right) {
            const mid = (left + right) >> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    };
    const l = search(target);
    const r = search(target + 1);
    return l === r ? [-1, -1] : [l, r - 1];
}

Rust

impl Solution {
    pub fn search_range(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let n = nums.len();
        let search = |x| {
            let mut left = 0;
            let mut right = n;
            while left < right {
                let mid = left + (right - left) / 2;
                if nums[mid] < x {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            left
        };
        let l = search(target);
        let r = search(target + 1);
        if l == r {
            return vec![-1, -1];
        }
        vec![l as i32, (r - 1) as i32]
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var searchRange = function (nums, target) {
    function search(x) {
        let left = 0,
            right = nums.length;
        while (left < right) {
            const mid = (left + right) >> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
    const l = search(target);
    const r = search(target + 1);
    return l == r ? [-1, -1] : [l, r - 1];
};

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @param Integer $target
     * @return Integer[]
     */
    function searchRange($nums, $target) {
        $search = function ($x) use ($nums) {
            $left = 0;
            $right = count($nums);
            while ($left < $right) {
                $mid = intdiv($left + $right, 2);
                if ($nums[$mid] >= $x) {
                    $right = $mid;
                } else {
                    $left = $mid + 1;
                }
            }
            return $left;
        };

        $l = $search($target);
        $r = $search($target + 1);
        return $l === $r ? [-1, -1] : [$l, $r - 1];
    }
}

Kotlin

class Solution {
    fun searchRange(nums: IntArray, target: Int): IntArray {
        val left = this.search(nums, target)
        val right = this.search(nums, target + 1)
        return if (left == right) intArrayOf(-1, -1) else intArrayOf(left, right - 1)
    }

    private fun search(nums: IntArray, target: Int): Int {
        var left = 0
        var right = nums.size
        while (left < right) {
            val middle = (left + right) / 2
            if (nums[middle] < target) {
                left = middle + 1
            } else {
                right = middle
            }
        }
        return left
    }
}