LeetCode TwoSum + 回文数

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TwoSum这题就直接把数字和位置放进Map里，然后用TARGET减当前的数，用结果去Map里找就行。不算是好的解法。

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i< nums.length; i++) {
            if (map.containsKey(target - nums[i])) {
                result[0] = map.get(target - nums[i]);
                result[1] = i;
            }
            map.put(nums[i], i);
        }
        return result;
    }
}

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数字回文这道题，之前是做过类似的，拿链表测字符串回文的，这次要做的是不依赖成字符串和链表来检测数字的回文。也比较简单。

class Solution {
    public static boolean isPalindrome(int x) {
        if (x < 0) {
            return false;
        }
        if (x <10) {
            return true;
        }
        int leftPointer = 0;
        int rightPointer = 0;
        int remainder = x;
        while (remainder >= 10) {
            remainder = remainder / 10;
            rightPointer++;
        }
        while (leftPointer < rightPointer) {
            int leftDigit = getLeftDigit(x, rightPointer, leftPointer);
            int rightDigit = getRightDigit(x);
            if (leftDigit != rightDigit) {
                return false;
            }
            x = removeFirstAndLastDigit(x, rightPointer, leftPointer);
            leftPointer++;
            rightPointer--;
        }
        return true;
    }

    private static int removeFirstAndLastDigit(int x, int rightPointer, int leftPointer) {
        return x % (int)Math.pow(10, rightPointer - leftPointer) / 10;
    }

    private static int getLeftDigit(int x, int rightPointer, int leftPointer) {
        return x / (int)Math.pow(10, getCurrentLength(rightPointer, leftPointer));
    }

    private static int getRightDigit(int x) {
        return x % 10;
    }

    private static int getCurrentLength(int right, int left) {
        return right - left;
    }
}