60. Search Insert Position 【easy】
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.
Example
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0
Challenge
O(log(n)) time
解法一：
class Solution {
/**
* param A : an integer sorted array
* param target : an integer to be inserted
* return : an integer
*/
public:
int searchInsert(vector<int> &A, int target) {
if (A.size() == ) {
return ;
} int start = ;
int end = A.size() - ; while (start + < end) {
int mid = start + (end - start) / ; if (A[mid] == target) {
return mid;
}
else if (A[mid] < target) {
start = mid;
}
else if (A[mid] > target) {
end = mid;
}
} if (target <= A[start]) {
return start;
}
else if (target <= A[end]) {
return end;
}
else if (target > A[end]) {
return end + ;
}
}
};
最后返回的时候要格外注意，按照模板来说，最后返回的start或者end中是有可能等于我们要找的数的，如果都没有找到后面还要补刀一个-1；对应这个题不用补刀，因为是要找插入的位置，必定会有一个合理的插入位置。
解法二：
public class Solution {
public int searchInsert(int[] A, int target) {
if (A == null || A.length == ) {
return ;
}
int start = ;
int end = A.length - ;
int mid; if (target < A[]) {
return ;
}
// find the last number less than target
while (start + < end) {
mid = start + (end - start) / ;
if (A[mid] == target) {
return mid;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
} if (A[end] == target) {
return end;
}
if (A[end] < target) {
return end + ;
}
if (A[start] == target) {
return start;
}
return start + ;
}
}
解法三：
class Solution {
/**
* param A : an integer sorted array
* param target : an integer to be inserted
* return : an integer
*/
public:
int searchInsert(vector<int> &A, int target) {
// find first position >= target
if (A.size() == ) {
return ;
} int start = , end = A.size() - ;
while (start + < end) {
int mid = (end - start) / + start;
if (A[mid] >= target) {
end = mid;
} else {
start = mid;
}
} if (A[start] >= target) {
return start;
}
if (A[end] >= target) {
return end;
} return A.size();
}
};
大神解法，非常简便！
