leetcode刷题（28）——34. 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]

1.不符合题目要求的正确解法：
直接硬解，也就是线性解法，但是时间复杂度不符合O(log n)要求

2.分别计算出左边界和右边界

class Solution {
 
    public int[] searchRange(int[] nums, int target) {
        int leftIndex = findLeftRange(nums,target);
        int rightIndex = findRightRange(nums,target);
        return new int[]{leftIndex,rightIndex};
    }

    public int findLeftRange(int[] nums, int target){
        int L = 0;
        int R = nums.length;
        while(L<R){
            int mid = (L+R)/2;
            if(nums[mid]==target){
                R=mid;
            }else if(nums[mid]>target){
                R=mid;
            }else if(nums[mid]<target){
                L=mid+1;
            }
        }
        if(L==nums.length){
            return -1;
        }else if(nums[L]==target){
            return L;
        }else{
            return -1;
        }
    }

    public int findRightRange(int[] nums, int target){
        int L = 0;
        int R = nums.length;
        while(L<R){
            int mid = (L+R)/2;
            if(nums[mid]==target){
                L=mid+1;
            }else if(nums[mid]>target){
                R=mid;
            }else if(nums[mid]<target){
                L=mid+1;
            }
        }
        if(L==0){
            return -1;
        }else if(nums[L-1]==target){
            return L-1;
        }else{
            return -1;
        }
    }
}

这里有一篇很好的解释二分算法的文章，可以参考：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/er-fen-cha-zhao-suan-fa-xi-jie-xiang-jie-by-labula/

3.官方解法
官方解法其实和上面的找出左右界限是一样的，只是进行了代码整合

class Solution {
    // returns leftmost (or rightmost) index at which `target` should be
    // inserted in sorted array `nums` via binary search.
    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;
    }
}