题目:
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0
class Solution {
public:
int searchInsert(int A[], int n, int target) {
int ans;
ans = binarySearch(A, 0, n-1, target, n);
return ans;
}
private:
int binarySearch(int A[], int begin, int end, int target, int n) {
while(begin <= end) {
int mid = begin + (end - begin)/2;
if(target == A[mid])
return mid;
else if(target > A[mid]) {
if(mid+1 < n && target < A[mid+1])
return mid+1;
begin = mid + 1;
}
else {
if(mid-1 >= 0 && target > A[mid-1])
return mid;
end = mid -1;
}
}
if(begin == n)
return n;
if(end == -1)
return 0;
}
};
本文深入探讨了在已排序数组中查找目标值的正确位置,包括实现细节与常见实例解析。
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