LeetCode Review Diary: 34. Find the first and last position of an element in a sorted array

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

方法1：

class Solution {
    
    public int[] searchRange(int[] nums, int target) {
    
        int start = -1, end = -1;
        int i = 0, j = nums.length-1;
        if(nums.length == 0 || nums == null){
    
            return new int[]{
    start, end};
        }
        while (i <= j){
    
            if(target != nums[i]){
    
                i++;
            }else if(target != nums[j]){
    
                j--;
            }else if(target == nums[i] && target == nums[j]){
    
                start = i;
                end = j;
                break;
            }
        }
        return new int[]{
    start,end};
    }
}

方法2：

class Solution {
    
    public int[] searchRange(int[] nums, int target) {
    
        int start = -1, end = -1;
        int i = 0, j = nums.length-1;
        while (i <= j){
    
            if(target != nums[i]){
    
                i++;
            }else if(target != nums[j]){
    
                j--;
            }else if(target == nums[i] && target == nums[j]){
    
                start = i;
                end = j;
                break;
            }
        }
        return new int[]{
    start,end};
    }
}

方法3：

class Solution {
    
    public int[] searchRange(int[] nums, int target) {
    
         int a = -1,b=-1,c=0;
        for(int i=0;i<nums.length;i++){
    
            if(nums[i]==target){
    
               if(c==0){
    
                   a = i;
                   c = 1;
               }
               b = i;
            }
        }
        return new int[] {
    a,b};
    }
}

官方题解：二分算法

class Solution {
    
    public int[] searchRange(int[] nums, int target) {
    
        int leftIdx = binarySearch(nums, target, true);
        int rightIdx = binarySearch(nums, target, false) - 1;
        if (leftIdx <= rightIdx && rightIdx < nums.length && nums[leftIdx] == target && nums[rightIdx] == target) {
    
            return new int[]{
    leftIdx, rightIdx};
        } 
        return new int[]{
    -1, -1};
    }

    public int binarySearch(int[] nums, int target, boolean lower) {
    
        int left = 0, right = nums.length - 1, ans = nums.length;
        while (left <= right) {
    
            int mid = (left + right) / 2;
            if (nums[mid] > target || (lower && nums[mid] >= target)) {
    
                right = mid - 1;
                ans = mid;
            } else {
    
                left = mid + 1;
            }
        }
        return ans;
    }
}