34-find-first-and-last-position-of-element-in-sorted-array.md

34. Find First and Last Position of Element in Sorted Array - 在排序数组中查找元素的第一个和最后一个位置

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。

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值，返回 [-1, -1]。

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]

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题目标签：Array / Binary Search

题目链接：LeetCode / LeetCode中国

题解

考查二分法及其变体。

二分法比较简单，然而，并不容易写对。主要是下标越界、死循环、是否取等号、+1 -1、变left还是right等问题。。

Language	Runtime	Memory
java	0 ms	44.4 MB

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if (nums.length == 0) return new int[]{-1, -1};
        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 st = left;
        left = st; 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[]{st, left};
    }
}

Language	Runtime	Memory
cpp	12 ms	N/A

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while(left <= right){
            int mid = (left + right) / 2;
            if(nums.at(mid) == target){
                int lleft = left, lright = right;
                while(lleft <= lright){
                    int lmid = (lleft + lright) / 2;
                    if(nums.at(lmid) >= target){
                        lright = lmid - 1;
                    }else{
                        lleft = lmid + 1;
                    }
                }
                int rleft = left, rright = right;
                while(rleft <= rright){
                    int rmid = (rleft + rright) / 2;
                    if(nums.at(rmid) > target){
                        rright = rmid - 1;
                    }else{
                        rleft = rmid + 1;
                    }
                }
                return vector<int>{lleft, rright};
            }else if(nums.at(mid) > target){
                right = mid - 1;
            }else{
                left = mid + 1;
            }
        }
        return vector<int>{-1, -1};
    }
};