679.24點遊戲

題解

class Solution {

    public boolean judgePoint24(int[] nums) {
        return backTrack(nums, 0);
    }

    // 第一步：求出所有排列，一一驗證
    public boolean backTrack(int[] nums, int index) {
        if (index == 4) {
            return judge(nums[0], nums[1], nums[2], nums[3]);
        }
        for (int i = index; i < 4; i++) {
            swap(nums, index, i);
            if (backTrack(nums, index + 1)) return true;
            swap(nums, index, i);
        }
        return false;
    }

    public void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }

    // 第二步：由于已經全排列，a、b、c、d 都是等價的，利用四種運算符選出三個數繼續
    public boolean judge(double a, double b, double c, double d) {
        return judge(a + b, c, d) ||
               judge(a - b, c, d) ||
               judge(a * b, c, d) ||
               judge(a / b, c, d);
    }

    // 第三步：a 是有兩個數組成的，b、c 隻表示一個數，等價
    public boolean judge(double a, double b, double c) {
        // 情況一：a 和 b(c) 組合，a 和 b(c) 不等價，- 号和 / 号需要考慮兩種情況
        return judge(a + b, c) ||
               judge(a - b, c) ||
               judge(a * b, c) ||
               judge(a / b, c) ||
               judge(b - a, c) ||
               judge(b / a, c) ||
               // 情況二：b 和 c 組合
               judge(a, b - c) ||
               judge(a, b + c) ||
               judge(a, b * c) ||
               judge(a, b / c);
    }

    // 第四步：a 和 b 不等價
    public boolean judge(double a, double b) {
        return Math.abs(a + b - 24) < 0.001 ||
               Math.abs(a - b - 24) < 0.001 ||
               Math.abs(a * b - 24) < 0.001 ||
               Math.abs(a / b - 24) < 0.001 ||
               Math.abs(b - a - 24) < 0.001 ||
               Math.abs(b / a - 24) < 0.001;
    }
}