二分查找

在一个有序的数组中，用二分查找来寻找目标元素。时间复杂度O(logn)

1、普通版二分查找

不考虑重复元素的情况

int binarySearch(vector<int>& data, int target) {    int left = 0, right = data.size() - 1;//注意“-1”    while (left <= right) {//注意“<=”        int mid = left + (right - left) / 2;//注意防止“溢出”        if (data[mid] < target)            mid = left + 1;        else if (data[mid] > target)            mid = right - 1;        else            return mid;    }    return -1;}

2、推广-有重复元素

2.1 元素第一次出现的位置

int binarySearch(vector<int>& data, int target) {    int left = 0, right = data.size() - 1;    while (left <= right) {        int mid = left + (right - left) / 2;        if (data[mid] < target)            left = mid + 1;        else            right = mid - 1;    }    if(left<data.size()&&data[left]==target)        return left;    return -1;}

测试用例：
{1, 1, 2, 3, 4},1) //0
{1, 2, 3, 4, 4}, 4) //3
{1, 2, 3, 3, 4}, 3) //2
{1, 2, 3, 4, 4}, 0) //-1
{1, 2, 3, 4, 4}, 5) //-1
{1, 1, 1, 1, 1}, 1) //0

2.2 元素最后一次出现的位置

int binarySearch(vector<int>& data, int target) {    int left = 0, right = data.size() - 1;    while (left <= right) {        int mid = left + (right - left) / 2;        if (data[mid] <= target)            left = mid + 1;        else            right = mid - 1;    }    if (right >= 0 && data[right] == target)        return right;    return -1;}

测试用例：
{1, 1, 2, 3, 4},1) //1
{1, 2, 3, 4, 4}, 4) //4
{1, 2, 3, 3, 4}, 3) //3
{1, 2, 3, 4, 4}, 0) //-1
{1, 2, 3, 4, 4}, 5) //-1
{1, 1, 1, 1, 1}, 1) //4

2.3第一个大于等于目标值

int binarySearch(vector<int>& data, int target) {    int left = 0, right = data.size() - 1;    while (left <= right) {        int mid = left + (right - left) / 2;        if (data[mid] < target)            left = mid + 1;        else            right = mid - 1;    }    return left;}

3.4最后一个小于等于目标值

int binarySearch(vector<int>& data, int target) {    int left = 0, right = data.size() - 1;    while (left <= right) {        int mid = left + (right - left) / 2;        if (data[mid] <= target)            left = mid + 1;        else            right = mid - 1;    }    return right;}

STL

low_bounder
Returns an iterator pointing to the first element in the range [first,last) which does not compare less than val.
第一个大于等于val的位置
upper_bounder
Returns an iterator pointing to the first element in the range [first,last) which compares greater than val.
第一个大于val的位置