Leetcode——在排序数组中查找元素的第一个和最后一个位置

Posted 2021-09-07 Yawn,

1. 在排序数组中查找元素的第一个和最后一个位置

跟在排序数组中查找数字 I（二分查找）.解法基本一致

（1）二分（找双边界）

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n = nums.length;
        int a = -1, b = -1;     //分别存储左右边界
        boolean flag = false;

        //寻找左边界
        int left = 0, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2 ;
            if (nums[mid] == target) {
                flag = true;
                right = mid;
            }
            else if (nums[mid] < target) 
                left = mid + 1;
            else  if (nums[mid] > target)
                right = mid;
        }
        a = left;

        //寻找右边界
        left = 0;
        right = n;
        while (left < right) {
            int mid = left + (right - left) / 2 ;
            if (nums[mid] == target) {
                flag = true;
                left = mid + 1;
            }
                
            else if (nums[mid] < target) 
                left = mid + 1;
            else  if (nums[mid] > target)
                right = mid;
        }
        //这里的right指目标值最后分割点，寻找到目标值后left = mid + 1; 所以需要减1
        b = left - 1;		

        if (!flag)
            return new int[]{-1, -1};
        return new int[]{a, b};

    }
}

（2）二分（找右边界）

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n = nums.length;
        int a = -1, b = -1;     //分别存储左右边界
        boolean flag = false;

        //寻找右边界
        int left = 0, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2 ;
            if (nums[mid] == target) {
                flag = true;
                left = mid + 1;
            }
                
            else if (nums[mid] < target) 
                left = mid + 1;
            else  if (nums[mid] > target)
                right = mid;
        }

        //这里的right指目标值最后分割点，寻找到目标值后left = mid + 1;	
        right = left - 1; 

        //找左边界
        left = right;   
        while (left >= 0 && nums[left] == target) {       //判断right > 0 即可
            left--;
        }
        if (!flag)
            return new int[]{-1, -1};
        return new int[]{left + 1, right};      //找左边界时，left--不满足>0 才返回，所以需要+1

    }
}

（3）二分（找左边界）

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int n = nums.length;
        int a = -1, b = -1;     //分别存储左右边界
        boolean flag = false;

        //寻找左边界
        int left = 0, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                flag = true;
                right = mid;
                //区间为[0,n) 左闭右开：nums[mid] == target，下一个值为mid + 1
                //[0,mid)，不包括mid
            } 
            else if (nums[mid] > target)
                right = mid;
            else if (nums[mid] < target)
                left = mid + 1;
        }

        //找左边界
        right = left;  
        while (right < n && nums[right] == target) {       //判断right > 0 即可
            right++;
        }
        if (!flag)
            return new int[]{-1, -1};
        return new int[]{left, right - 1};      //找右边界时，right++不满足< n 才返回，所以需要-1
    }
}