Search Insert Position

Source

• lintcode: (60) 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.

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

题解

应该把二分法的问题拆解为find the first/last position of...的问题。由最原始的二分查找可找到不小于目标整数的最小下标。返回此下标即可。

Java

public class Solution {
    /**
     * param A : an integer sorted array
     * param target :  an integer to be inserted
     * return : an integer
     */
    public int searchInsert(int[] A, int target) {
        if (A == null) {
            return -1;
        }
        if (A.length == 0) {
            return 0;
        }

        int start = 0, end = A.length - 1;
        int mid;

        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return mid; // no duplicates, if not `end = target;`
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }

        if (A[start] >= target) {
            return start;
        } else if (A[end] >= target) {
            return end; // in most cases
        } else {
            return end + 1; // A[end] < target;
        }
    }
}

源码分析

要注意例子中的第三个, [1,3,5,6], 7 → 4，即找不到要找的数字的情况，此时应返回数组长度，即代码中最后一个else的赋值语句return end + 1;