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Given a sorted array of integers, find the starting and ending position of a given target value
Your algorithm's runtime complexity must be in the order of O(log n).
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].
解题分析：就是一个二分搜索的应用，不难。但是就是注意一个数组越界的检查。LeetCode自己的编译器会报错。
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int a,b;
?a=num;
?b=num;
while(A[a] == target ) a--;
while(A[b] == target ) b++;
最开始的时候，代码是这样的，对二分搜索找到了target，再进行往前往后的找。在codeblocks上运行是无错， 但就是提交在LeetCode上面会出现问题。
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最后做了一个数组的溢出判断，终于AC了
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int a,b;
a=num;
b=num;
while(A[a] == target && a >= 0) a--;
while(A[b] == target && b <= n-1) b++;
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class Solution {
private: int binary_search(int* a, int len, int goal)
{
int low = 0;
int high = len - 1;
while(low <= high)
{
int middle = (low + high)/2;
if(a[middle] == goal)
return middle;
else if(a[middle] > goal)
high = middle - 1;
else
low = middle + 1;
}
return -1;
}
public:
vector
searchRange(int A[], int n, int target) {
vector
s; int num = binary_search(A,n,target); if(num == -1) {s.push_back(-1);s.push_back(-1);} else { int a,b; a=num; b=num; while(A[a] == target && a >= 0) a--; while(A[b] == target && b <= n-1) b++; s.push_back((++a)); s.push_back((--b)); } return s; } };
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