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题目
给定一个包含 n 个整数的排序数组,找出给定目标值 target 的起始和结束位置。
如果目标值不在数组中,则返回[-1, -1]样例
给出[5, 7, 7, 8, 8, 10]和目标值target=8,
返回[3, 4]挑战
时间复杂度 O(log n)
题解
1.二分法找到target,再向两边查找边界。
public class Solution {
/**
*@param A : an integer sorted array
*@param target : an integer to be inserted
*return : a list of length 2, [index1, index2]
*/
public int[] searchRange(int[] A, int target) {
if (A.length == 1)
{
return A[0] == target ? new int[]{0,0} : new int[]{-1,-1};
}
return doSearch(A,0,A.length-1,target);
}
private int[] doSearch(int[] A, int start, int end, int target)
{
while (start <= end)
{
int mid = start + (end - start) / 2;
if (A[mid] == target)
{
int left=mid;
int right=mid;
while (left > 0 && A[left-1] == target)
{
left--;
}
while (right < A.length-1 && A[right+1] == target)
{
right++;
}
return new int[]{left,right};
}
else if (A[mid] < target)
{
start = mid + 1;
}
else
{
end = mid - 1;
}
}
return new int[]{-1,-1};
}
}
2.二分法分别查找下界和上界
public class Solution {
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 if (A[mid] < target) {
start = mid;
} else {
end = 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 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;
}
}
Last Update 2016.10.6