lcp329

交替数字和

方法一：模拟

class Solution {
    public int alternateDigitSum(int n) {
        String str = String.valueOf(n);
        int len = str.length();
        int res = 0;
        boolean flag = len % 2 == 1? true : false;
        for (int i = 0; i < len; ++i) {
            int x = n % 10;
            n /= 10;
            if (flag)
                res += x;
            else
                res -= x;
            flag = !flag;
        }

        return res;
    }
}

根据第 K 场考试的分数排序

方法一：排序

class Solution {
    public int[][] sortTheStudents(int[][] score, int k) {
        Arrays.sort(score, (o1, o2) -> o2[k] - o1[k]);
        return score;
    }
}

执行逐位运算使字符串相等

方法一：贪心

class Solution {
    public boolean makeStringsEqual(String s, String target) {
        int n = s.length();
        int cntS = 0, cntT = 0;
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == '1') ++cntS;
            if (target.charAt(i) == '1') ++ cntT;
        }
        if (cntT == 0)
            return cntS == 0 ? true : false;
        else
            return cntS > 0 ? true : false;
    }
}

拆分数组的最小代价

方法一：DP

class Solution {
    public int minCost(int[] nums, int k) {
        int n = nums.length;
        int[] f = new int[n + 1];
        Arrays.fill(f, Integer.MAX_VALUE >> 1);
        f[0] = 0;
        for (int i = 1; i <= n; ++i) {
            int[] cnt = new int[n];
            int t = 0;
            for (int j = i - 1; j >= 0; --j) {
                int x = ++cnt[nums[j]];
                if (x == 2) t += 2;
                else if (x > 2) t += 1;
                f[i] = Math.min(f[i], t + f[j] + k);
            }
        }
        return f[n];
    }
}