[数据结构与算法]Java动态规划

剑指offer 63.股票的最大利润

class Solution {
    public int maxProfit(int[] prices) {
        int buy = Integer.MAX_VALUE, maxProfit = 0;
        for(int price : prices){
            buy = Math.min(price, buy);
            maxProfit = Math.max(maxProfit, price - buy);
        }
        return maxProfit;
    }
}

剑指offer 42.连续子数组的最大和

class Solution {
    public int maxSubArray(int[] nums) {
        int res = nums[0];
        for(int i = 1; i < nums.length; i++){
            nums[i] += Math.max(nums[i - 1], 0);
            res = Math.max(nums[i], res);
        }
        return res;
    }
}

剑指offer 47.礼物的最大价值

class Solution {
    public int maxValue(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(i == 0 && j == 0) continue;
                if(i == 0) grid[i][j] += grid[i][j - 1];
                else if(j == 0) grid[i][j] += grid[i - 1][j]; 
                else grid[i][j] += Math.max(grid[i - 1][j], grid[i][j - 1]);
            }
        }
        return grid[m - 1][n - 1];
    }
}

剑指offer 46.把数字翻译成字符串

解题思路:

compareTo()方法

tmp.compareTo("10"): 如果tmp > 10, 则返回一个大于0的值

tmp.compareTo("25"): 如果tmp < 25, 则返回一个小于0的值

tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0: 说明tmp的值在[10, 25]区间之内

class Solution {
    public int translateNum(int num) {
        String str = String.valueOf(num);
        int[] dp = new int[str.length() + 1];
        dp[0] = dp[1] = 1;
        for(int i = 2; i <= str.length(); i++){
            String tmp = str.substring(i - 2, i);
            if(tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0){
                dp[i] = dp[i - 2] + dp[i - 1];
            }else{
                dp[i] = dp[i - 1];
            }
        }
        return dp[str.length()];
    }
}

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