Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order ofO(logn).
If the target is not found in the array, return[-1, -1].
For example,
Given[5, 7, 7, 8, 8, 10]and target value 8,
return[3, 4].
題目簡述：
給定一個升序排列的int 數(shù)組，找到給定數(shù)值的起始和終止位置挨决。
時間復(fù)雜度要求 O（logn）
復(fù)雜度要求logn 首先考慮的就是binary search鬓长。
復(fù)雜度為O（N2）的思路很簡單蚂四，計算nums[i] == target 的個數(shù)踏幻，主要考慮一下找不到的情況谱醇，然后將數(shù)組賦值為[ -1, -1 ]漱牵。
JAVA 版本：
public int[] searchRange(int[] A, int target) {
if (A.length == 0) {
return new int[]{-1, -1};
}
int start, end, mid;
int[] bound = new int[2];
// search for left bound
start = 0;
end = A.length - 1;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (A[mid] >= target) {
end = mid;
} else {
start = mid;
}
}
if (A[start] == target) {
bound[0] = start;
} else if (A[end] == target) {
bound[0] = end;
} else {
bound[0] = bound[1] = -1;
return bound;
}
// search for right bound
start = 0;
end = A.length - 1;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (A[mid] <= target) {
start = mid;
} else {
end = mid;
}
}
if (A[end] == target) {
bound[1] = end;
} else if (A[start] == target) {
bound[1] = start;
} else {
bound[0] = bound[1] = -1;
return bound;
}
return bound;
}
