034在排序数组中查找元素的第一个和最后一个

#include "000库函数.h"

//使用二分法查找到目标值的位置，然后分两边再查找出起始位置和终止位置
//16ms  不是严格意义上的logn的复杂度
class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int>v = { -1,-1 };
        int i = 0, j = nums.size() - 1;
        int m = -1;
        while (i <= j) {
            int t = (i + j) / 2;
            if (nums[t] == target) {
                m = t;
                break;
            }
            if (nums[t] > target)    j = t - 1;
            else i = t + 1;
        }
        if (m == -1) return v;    
        int t = m;
        while (t >= 0 && nums[t] == target) --t; 
        v[0] = (t + 1);
        t = m;
        while (t < nums.size() && nums[t] == target) ++t;
        v[1] = (t - 1);
        return v;
    }
};

//复杂度满足logn   
class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {        
        vector<int> res(2, -1);
        if (nums.size() == 0)return res;
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) left = mid + 1;//使用这种判断，会找到最左边的目标值
            else right = mid;
        }
        if (nums[right] != target) return res;

        res[0] = right;
        right = nums.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] <= target) left = mid + 1;//使用这种判断，会找到最右边的目标值
            else right = mid;
        }
        res[1] = left - 1;
        return res;
    }
};
void T034() {
    Solution s;
    vector<int >n;
    n = { 8 };
    n = s.searchRange(n, 8);
    cout << n[0] << "  " << n[1] << endl;
    n = { 5,7,7,8,8,10 };
    n = s.searchRange(n, 9);
    cout << n[0] << "  " << n[1] << endl;
    n = { 5,7,7,8,8,8,10 };
    n = s.searchRange(n, 8);
    cout << n[0] << "  " << n[1] << endl;
    n = { 1,1,1,1,1,1};
    n = s.searchRange(n,1);
    cout << n[0] << "  " << n[1] << endl;

}