动态规划字符串

题目

https://leetcode-cn.com/problems/ba-shu-zi-fan-yi-cheng-zi-fu-chuan-lcof

解法

从上至下dp[i] 递归分析，从下至上循环解题

public int translate(int num) {
        if (num < 0) return -1;
        String src = String.valueOf(num);
        int len = src.length();
        int[] dp = new int[len];
//        dp[-1] = 0;
        dp[0] = 1;

        for (int i=1; i<len; i++) {
//            dp[i] = dp[i-1] + dp[i-2];
            dp[i] += dp[i-1];

            String str = src.substring(i-1, i+1);
            if (str.compareTo("10") >=0 && str.compareTo("25") <=0) {
                if (i == 1) {
                    dp[i] += 1;
                    continue;
                }
                dp[i] += dp[i-2];
            }
        }
        return dp[len-1];
    }

2020-07-06

LC63_uniquePathsWithObstacles

动态规划字符串

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