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界面组件DevExpress Reporting v24.1预览版 - 拥有原生Angular报表查看器

三分梦_0bc3 2024-05-15 阅读 6

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java的Deque

Leetcode 1143. 最长公共子序列

解题思路

和上一题的区别是,不初始化0,如果没匹配到就变为左边数

代码

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int res = 0;
        int [][]dp = new int[text1.length()+1][text2.length()+1];
        for (int i = 1; i <= text1.length(); i++) {
            for (int j = 1; j <= text2.length(); j++) {
                if (text1.charAt(i-1) == text2.charAt(j-1))
                    dp[i][j] = dp[i-1][j-1] + 1;
                else
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                res = Math.max(res, dp[i][j]);
            }
        }
        return res;
    }
}

//滚动数组
class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int res = 0;
        int []dp = new int[text2.length()+1];
        for (int i = 1; i <= text1.length(); i++) {
            for (int j = text2.length(); j > 0; j--) {
                if (text1.charAt(i-1) == text2.charAt(j-1))
                    dp[j] = dp[j-1] + 1;
                else
                    dp[j] = Math.max(dp[j], dp[j-1]);
                res = Math.max(res, dp[j]);
            }
        }
        return res;
    }
}

总结

暂无

Leetcode 1035. 不相交的线

解题思路

上秒那题包了个马甲

代码

class Solution {
    public int maxUncrossedLines(int[] nums1, int[] nums2) {
        int res = 0;
        int [][]dp = new int[nums1.length+1][nums2.length+1];
        for (int i = 1; i <= nums1.length; i++) {
            for (int j = 1; j <= nums2.length; j++) {
                if (nums1[i-1] == nums2[j-1])
                    dp[i][j] = dp[i-1][j-1] + 1;
                else
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                res = Math.max(res, dp[i][j]);
            }
        }
        return res;
    }
}

总结

暂无

Leetcode 53. 最大子数组和

解题思路

dp数组的含义是到这位置的最大和是多少

代码

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

总结

暂无

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