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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

两遍二分，分别查找出左边界和右边界。

以下是二分查找的两个通用模板：

模板 1：

boolean check(int x) {}

int search(int left, int right) {
    while (left < right) {
        int mid = (left + right) >> 1;
        if (check(mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

模板 2：

boolean check(int x) {}

int search(int left, int right) {
    while (left < right) {
        int mid = (left + right + 1) >> 1;
        if (check(mid)) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return left;
}

做二分题目时，可以按照以下步骤：

  1. 写出循环条件：while (left < right)，注意是 left < right，而非 left <= right；
  2. 循环体内，先无脑写出 mid = (left + right) >> 1；
  3. 根据具体题目，实现 check() 函数（有时很简单的逻辑，可以不定义 check），想一下究竟要用 right = mid（模板 1） 还是 left = mid（模板 2）；
    • 如果 right = mid，那么无脑写出 else 语句 left = mid + 1，并且不需要更改 mid 的计算，即保持 mid = (left + right) >> 1；
    • 如果 left = mid，那么无脑写出 else 语句 right = mid - 1，并且在 mid 计算时补充 +1，即 mid = (left + right + 1) >> 1。
  4. 循环结束时，left 与 right 相等。

注意，这两个模板的优点是始终保持答案位于二分区间内，二分结束条件对应的值恰好在答案所处的位置。 对于可能无解的情况，只要判断二分结束后的 left 或者 right 是否满足题意即可。

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 == len(nums) or 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 == nums.length || l >= r ? new int[]{-1, -1} : new int[]{l, r - 1};
    }

    private int search(int[] nums, int target) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] >= target) {
                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 == nums.size() || l >= r) return {-1, -1};
        return {l, r - 1};
    }
};

JavaScript

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

Go

func searchRange(nums []int, target int) []int {
	search := func(target int) int {
		left, right := 0, len(nums)
		for left < right {
			mid := (left + right) >> 1
			if nums[mid] >= target {
				right = mid
			} else {
				left = mid + 1
			}
		}
		return left
	}
	l, r := search(target), search(target+1)
	if l == len(nums) || l >= r {
		return []int{-1, -1}
	}
	return []int{l, r - 1}
}

Rust

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

TypeScript

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

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