二分查找

1. 找>=target的第一个位置
2. 找<=target的最后一个位置

class Solution {
public:
    //找>=target的第一个位置
    int binarySearchLeft(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] >= target) {
                right = mid - 1;//意味着索引>=right+1的数组值均>=target
            }
            else {
                left = mid + 1;//意味着索引<=left-1的数组值均<target
            }
        }
        return left;//左闭右闭区间退出循环，有left=right+1
    }
    //找<=target的最后一个位置
    int binarySearchRight(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] <= target) {
                left = mid + 1;//意味着索引<=left-1的数组值均<=target
            }
            else {
                right = mid - 1;//意味着索引>=right+1的数组值均>target
            }

        }
        return right;//左闭右闭区间退出循环，有left=right+1
    }
    vector<int> searchRange(vector<int>& nums, int target) {
        int leftIdx = binarySearchLeft(nums, target);
        int rightIdx = binarySearchRight(nums, target);
        if (leftIdx <= rightIdx && rightIdx < nums.size() && nums[leftIdx] == target && nums[rightIdx] == target) {
            return vector<int>{leftIdx, rightIdx};
        }
        return vector<int>{-1, -1};
    }
};

力扣 34. 在排序数组中查找元素的第一个位置和最后一个位置