README.md

面试题 53 - I. 在排序数组中查找数字 I

题目描述

统计一个数字在排序数组中出现的次数。

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: 2

示例  2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: 0

限制：

• 0 <= 数组长度 <= 50000

解法

两遍二分，分别查找出左边界和右边界。

Python3

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if len(nums) == 0:
            return 0
        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 0
        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 left - l + 1

Java

class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 0) {
            return 0;
        }
        // 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 0;
        }
        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 left - l + 1;
    }
}

C++

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        int left = 0, right = n;
        int first, last;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        if (left == n || nums[left] != target) {
            return 0;
        }
        first = left;
        left = 0, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        last = left - 1;
        return last - first + 1;
    }
};

Go

func search(nums []int, target int) int {
	if len(nums) == 0 {
		return 0
	}
	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 0
	}
	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 left - l + 1
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    if (nums.length == 0) {
        return 0;
    }
    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 0;
    }
    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 left - l + 1;
};

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