LeetCode刷题笔记 34

题目：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。
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]。

我的答案：
思路：二分法

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int low=0,high=nums.length-1,start=-1,end=-1;
        while(low<=high){
            int mid = (low + high)/2;
            if(nums[mid]>target)
                high = mid-1;
            else if(nums[mid]<target)
                low = mid+1;
            else{   //找到符合条件的位置之后，在该位置向左向右寻找起始和结束的位置
                int mid1=mid,mid2=mid;
                while(mid1>=0 && nums[mid1]==target){
                    mid1--;
                }
                while(mid2<nums.length && nums[mid2]==target){
                    mid2++;
                }
                start=mid1+1;
                end=mid2-1;
                return new int[]{start,end};
            }
        }
        return new int[]{start,end};
    }
}

答案：

class Solution {
 //二分查找，查找左索引时判断nums>=target，直至low=high；
 //查找右索引时判断nums>target，直至low=high
    private int extremeInsertionIndex(int[] nums, int target, boolean left) {
        int lo = 0;
        int hi = nums.length;

        while (lo < hi) {
            int mid = (lo + hi) / 2;
            if (nums[mid] > target || (left && target == nums[mid])) {
                hi = mid;
            }
            else {
                lo = mid+1;
            }
        }

        return lo;
    }

    public int[] searchRange(int[] nums, int target) {
        int[] targetRange = {-1, -1};

        int leftIdx = extremeInsertionIndex(nums, target, true);

        // assert that `leftIdx` is within the array bounds and that `target`
        // is actually in `nums`.
        if (leftIdx == nums.length || nums[leftIdx] != target) {
            return targetRange;
        }

        targetRange[0] = leftIdx;
        targetRange[1] = extremeInsertionIndex(nums, target, false)-1;

        return targetRange;
    }
}