34.Find First and Last Position of Element in Sorted Array---头条面试题、《剑指offer》38

题目链接

题目大意：找出一串升序数组中target值的起始下标和结束下标值，如果不存在则返回{-1，-1}。

解法一：用二分查找，找到数组中的target，然后找其左边和右边的target下标值。代码如下（耗时11ms）：

    public int[] searchRange(int[] nums, int target) {
        if(nums == null || nums.length == 0) {
            int[] r = {-1, -1};
            return r;
        }
        int low = 0, high = nums.length - 1;
        int start = -1, end = -1;
        while(low <= high) {
            int mid = (low + high) / 2;
            if(nums[mid] < target) {
                low = mid + 1;
            }
            else if(nums[mid] > target) {
                high = mid - 1;
            }
            else {
                //找左边起始下标
                for(int i = mid; i >= 0; i--) {
                    if(nums[i] == target) {
                        start = i;
                    }
                    else {
                        break;
                    }
                }
                //找右边终止下标
                for(int i = mid; i < nums.length; i++) {
                    if(nums[i] == target) {
                        end = i;
                    }
                    else {
                        break;
                    }
                }
                break;
            }
        }
        int[] res = {start, end};
        return res;
    }

解法二：直接暴力，一次遍历，找重复值。代码如下（耗时10ms）：

    public int[] searchRange(int[] nums, int target) {
        int start = -1, end = -1;
        boolean mark = false;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == target) {
                if(mark == false) {
                    start = end = i;
                    mark = true;
                }
                else {
                    end = i;
                }
            }
        }
        int[] res = {start, end};
        return res;
    }

 解法三：真正的二分查找。法一其实复杂度还是o(n)。应该先对起始下标进行二分查找，然后再对结束下标进行二分查找。代码如下（耗时5ms）：

    public int[] searchRange(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int[] res = {-1, -1};
        if(nums.length == 0) {
            return res;
        }
        //二分找到起始下标
        while(left < right) {
            int mid = (left + right) / 2;
            //这里比较左值，如果<，则left更新，否则left不会更新
            //所以left不更新有两种情况：>或=
            if(nums[mid] < target) {
                left = mid + 1;
            }
            //这里统统修改右值
            else {
                right = mid;
            }
        }
        if(nums[left] != target) {
            return res;
        }
        res[0] = left;
        left = 0;
        right = nums.length - 1;
        //二分找到结束下标
        while(left < right) {
            //这里要+1，否则会出错
            int mid = (left + right) / 2 + 1;
            //比较右值，如果>，则right更新
            if(nums[mid] > target) {
                right = mid - 1;
            }
            //修改Left
            else {
                left = mid;
            }
        }
        res[1] = right;
        return res;
    }