Java for LeetCode 035 Search Insert Position

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

解题思路一：

二分查找，注意边界条件，JAVA实现如下：

static public int searchInsert(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while (left <= right) {
if (target == nums[(left + right) / 2])
return (left + right) / 2;
else if (target < nums[(left + right) / 2]) {
if ((left + right) / 2 - 1 < left)
return target > nums[left] ? left + 1 : left;
if (target > nums[(left + right) / 2 - 1])
return (left + right) / 2;
if (target == nums[(left + right) / 2 - 1])
return (left + right) / 2 - 1;
right = (left + right) / 2 - 1;

} else {
if ((left + right) / 2 + 1 > right)
return target > nums[right] ? right + 1 : right;
if (target <= nums[(left + right) / 2 + 1])
return (left + right) / 2 + 1;
left = (left + right) / 2 + 1;
}
}
return right;
}

解题思路二：

同样采用二分，可以通过某些方法，减少代码量，JAVA实现如下：

static public int searchInsert(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while (left < right) {
if (nums[(right + left) / 2] < target)
left = (right + left) / 2 + 1;
else
right =  (right + left) / 2;
}
return nums[left] >= target ? left : left + 1;
}