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;
     }