//尋找val出現的位置,不存在時返回-1(數組不存在重複元素)
int binarySearch(vector<int> arr, int val) {
int l = 0, r = arr.size() - 1;
while (l <= r) {
int mid = (l+r)/2;
if (arr[mid] == val) {
return mid;
}
else if(arr[mid] < val) {
l = mid + 1;
}
else {
r = mid - 1;
}
}
return -1;
}
//如果存在val,返回val所在位置的索引(最左邊),否則返回小於val的元素最大索引
int floor(vector<int> arr, int val) {
int n = arr.size();
int l = 0, r = n - 1;
while (l < r) {
int mid = l + (r - l + 1) / 2;
if (arr[mid] >= val) {
r = mid - 1;
}
else {
l = mid;
}
}
if (l + 1 < n && arr[l + 1] == val) {
return l + 1;
}
return l;
}
//如果存在val,返回val所在位置的索引(最右邊),否則返回大於val的元素最小索引
int ceil(vector<int> arr, int val) {
int n = arr.size();
int l = 0, r = n - 1;
while (l < r) {
int mid = l + (r - l) / 2;
if (arr[mid] <= val) {
l = mid + 1;
}
else {
r = mid;
}
}
if (l - 1 > 0 && arr[l - 1] == val) {
return l - 1;
}
return l;
}