Leetcode 35. 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

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解题思路：

binary search 经典思路

四点要素:
1. start + 1 < end
2. start + (end - start) / 2

3. A[mid] ==, <, >

4. A[start] A[end] ? target 

把题目逻辑想清楚，其实就是找出多少数比target小的，也可以想成找第一个大于等于目标数的位置。

注意中间那块是模板。

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Java code:

1. find the last position < target 

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int start = 0;
        int end = nums.length - 1;
        int mid;
        
        if(target < nums[0]) {
            return 0;
        }
        
        //find the last number less than target
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                end = mid;  
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (nums[end] == target) {
            return end;
        }
        if(nums[end] < target) {
            return end + 1;
        }
        if(nums[start] == target) {
            return start;
        }
        return start + 1;
    }
}

2. find the first position >= target

public class Solution {
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int start = 0;
        int end = nums.length - 1;
        int mid;
        
        //find the first position >= target
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                end = mid;  
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (nums[start] >= target) {
            return start;
        }else if (nums[end] >= target) {
            return end;
        }else {
            return end+1;
        }
    }
}

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Reference:

1. http://www.jiuzhang.com/solutions/search-insert-position/