35. Search Insert Position

    /*
     * 35. Search Insert Position 
     * 2016-5-2 by Mingyang 这道题目偏简单
     * 这只是常规的解法，后面我们来logn的解法
     */
    public int searchInsert(int[] nums, int target) {
        if (nums == null || nums.length == 0)
            return 0;
        if (target < nums[0])
            return 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] >= target)
                return i;
        }
        return nums.length;
    }
    /*
     * 然后自己给出了一个Binary Search的解法，
     * 但是思路非常混乱 列下来以儆效尤
     */
    public int searchInsert1(int[] nums, int target) {
        int len = nums.length;
        if (nums == null || len == 0)
            return 0;
        if (target < nums[0])
            return 0;
        if (target > nums[len - 1])
            return len;
        int start = 0;
        int end = len - 1;
        while (start < end) {
            int mid = (start + end) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                start = mid;
                if (start + 1 == end)
                    return end;
            } else {
                end = mid;

            }
        }
        return start;
    }
    // 史上最标准答案，最简洁的一种BS的方法，只需要了解为什么后面返回low就好了
    public int searchInsert2(int[] nums, int target) {
        int low = 0, high = nums.length - 1;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (nums[mid] == target)
                return mid;
            else if (nums[mid] > target)
                high = mid - 1;
            else
                low = mid + 1;
        }
        return low;
    }