Search Insert Position

Question: (60) Search Insert Position

题解：

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。

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：已在源码处注释