一、朴素二分

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        int left = 0, right = nums.size();
        while (left <= right) {
            int mid = left + ((right - left) >> 1);  // 防止溢出
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return -1;

二、整数二分

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class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
    // 寻找左区间右端点，右区间左端点
    int l = 0, r = nums.size() - 1;
    vector<int> ans = {-1, -1};
    if (nums.size() == 0) return ans;
    while (l < r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] >= target) r = mid;
        else l = mid + 1;
    }

    if (nums[l] == target) ans[0] = l;
    else return ans;

    r = nums.size() - 1;
    while (l < r) {
        int mid = l + (r - l + 1) / 2;
        if (nums[mid] <= target) l = mid;
        else r = mid - 1;
    }
    ans[1] = l;
    return ans;
}
};