LeetCode解题报告（360) -- 34. Find First and Last Position of Element in Sorted Array

Problem

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

Follow up: Could you write an algorithm with O(log n) runtime complexity?

Example 1:

1 2	Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4]

Example 2:

1 2	Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1]

Example 3:

1 2	Input: nums = [], target = 0 Output: [-1,-1]

Constraints:

0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums is a non-decreasing array.
-109 <= target <= 109

Analysis

题目给出一个有序数组，还有一个target，要求找出target在数组中第一次和最后一次出现的下标。其实题目非常简单了，因为数组是有序的，所以也保证了相同的值是靠在一起的，从左往右开始遍历，我们只需要在第一次出现的时候记录下一个下标，然后最后一次出现的时候记录下标，这就是答案了。

Solution

无

Code

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int size = nums.size();
        vector<int> result;
        int length = 0;
        bool flag = true;
        for (int i = 0; i < size; i++) {
            if (nums[i] == target) {
                if (flag) {
                    flag = false;
                    result.push_back(i);                    
                }
                length++;
            }
        }
        if (flag) {
            result.push_back(-1);
            result.push_back(-1);
        } else {
            result.push_back(result[0] + length - 1);
        }
        return result;
    }
};

Summary

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