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  • [LeetCode] 34

    Given a sorted array of integers, find the starting and ending position of a given target value.

    Your algorithm's runtime complexity must be in the order of O(log n).

    If the target is not found in the array, return [-1, -1].

    For example,
    Given [5, 7, 7, 8, 8, 10] and target value 8,
    return [3, 4].

    class Solution {
    public:
      vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res({-1, -1});
        if (nums.empty()) {
          return res;
        }
        int start = 0;
        int end = nums.size() -1;
        int idx = -1;
        while (start <= end) {
          int mid = (end + start) >> 1; // /2
          if (nums[mid] == target) {
            idx = mid;
            break;
          }
          if (nums[mid] > target) {
            end = mid -1;
          }
          else {
            start = mid + 1;
          }
        }
        if (idx == -1) {
          return res;
        }
        for (end = idx; end < nums.size() -1; ++end) {
          if (nums[end] != nums[end + 1]) {
            break;
          }
        }
        for (start = idx; start >0 ; --start) {
          if (nums[start] != nums[start -1 ]) {
            break;
          }
        }
        res[0] = start;
        res[1] = end;
        return res;
      }
    };

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  • 原文地址:https://www.cnblogs.com/shoemaker/p/4772978.html
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