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
public class Solution {
public int searchInsert(int[] nums, int target) {
int len=nums.length;
for(int i =0;i<len;i++){
if(nums[i]==target){
return i;
}
else if(nums[i]<target&&i+1<len&&nums[i+1]>target){
return i+1;
}
else if(nums[len-1]<target){
return len;
}
else if(nums[0]>target){
return 0;
}
}
return 0;
}
}
解题思路：
- 判断边界值最小返回0最大返回数组长度，和某个值一样返回该值下标+1,在两个元素之间返回较大元素下标
- 注意在两个元素之间的时候判断防止溢出
// version 1: find the first position >= target
public class Solution {
public int searchInsert(int[] A, int target) {
if (A == null || A.length == 0) {
return 0;
}
int start = 0, end = A.length - 1; while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (A[mid] == target) {
return mid;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
} if (A[start] >= target) {
return start;
} else if (A[end] >= target) {
return end;
} else {
return end + 1;
}
}
} // version 2: find the last position < target, return +1， 要特判一下target小于所有数组里面的元素 public class Solution {
public int searchInsert(int[] A, int target) {
if (A == null || A.length == 0) {
return 0;
}
int start = 0;
int end = A.length - 1;
int mid; if (target < A[0]) {
return 0;
}
// find the last number less than target
while (start + 1 < end) {
mid = start + (end - start) / 2;
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 + 1;
}
if (A[start] == target) {
return start;
}
return start + 1;
}
}- 以上两个代码为两种更高效的方法