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

给定一个按照升序排列的整数数组 nums，和一个目标值 target。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]。

进阶：

• 你可以设计并实现时间复杂度为 O(log n) 的算法解决此问题吗？

 

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

二分查找。

两遍二分，分别查找出左边界和右边界。

Python3

class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        if len(nums) == 0:
            return [-1, -1]
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left + right) >> 1
            if nums[mid] >= target:
                right = mid
            else:
                left = mid + 1
        if nums[left] != target:
            return [-1, -1]
        l, right = left, len(nums) - 1
        while left < right:
            mid = (left + right + 1) >> 1
            if nums[mid] <= target:
                left = mid
            else:
                right = mid - 1
        return [l, left]

Java

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if (nums.length == 0) {
            return new int[]{-1, -1};
        }
        // find first position
        int left = 0, right = nums.length - 1;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        if (nums[left] != target) {
            return new int[]{-1, -1};
        }
        int l = left;

        // find last position
        right = nums.length - 1;
        while (left < right) {
            int mid = (left + right + 1) >>> 1;
            if (nums[mid] <= target) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return new int[]{l, left};
    }
}

C++

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if (nums.size() == 0) {
            return vector<int>{-1, -1};
        }
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + right >> 1;
            if (nums[mid] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        if (nums[left] != target) {
            return vector<int>{-1, -1};
        }
        int l = left;
        right = nums.size() - 1;
        while (left < right) {
            int mid = left + right + 1 >> 1;
            if (nums[mid] <= target) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return vector<int>{l, left};
    }
};

JavaScript

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

Go

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

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