題目:使用二分法查找循環有序數組的最小值。
int getSubScriptFromArray(int arr[], int left, int right){
while (left < right) {
int mid = left + (right - left) / 2;
if (arr[mid] >arr[right]) {
//左邊有序, 最小值在右邊,不包含mid
left = mid + 1;
} else if (arr[mid] < arr[right]) {
// 右邊有序, 最小值在左邊,可能是mid
right = mid;
} else{
right = right - 1;
}
}
return arr[left];
}
擴展一:
二分法查找循環有序數組中的某一元素,如果存在返回該值,否則返回 -1。
int getNumFromLooOrderArray(int arr[], int target, int left, int right) {
cout<<"len: "<< getArrayLen(arr) << endl;
while (left < right) {
int mid = left + (right - left) / 2;
cout<<"left: " << left << " right: " << right << endl;
if (arr[mid] == target) {
return arr[mid];
}
if (arr[mid] < arr[right]) {
if (target > arr[mid] && target < arr[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
if (target > arr[right]) {
right = mid - 1;
}
} else{
// mid 左邊有序
if (target < arr[mid] && target > arr[left]) {
right = mid - 1;
} else {
left = mid + 1;
}
}
}
if (target == arr[left]) {
return arr[left];
}
return -1;
}