1004. 最大连续1的个数 III

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该博客主要介绍了如何解决一个数组中，最多翻转K个0变为1，找到最长全1子数组的问题。作者提供了一个Java解决方案，通过遍历数组并维护两个指针来实现。算法思路是动态规划，但存在效率问题，因为它的时间复杂度较高，可能不适用于大规模数据。博客还提到，解题关键在于有效地减少不必要的计算。

给定一个由若干 0 和 1 组成的数组 A，我们最多可以将 K 个值从 0 变成 1 。

返回仅包含 1 的最长（连续）子数组的长度。

示例 1：

输入：A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
输出：6
解释： 
[1,1,1,0,0,1,1,1,1,1,1]
粗体数字从 0 翻转到 1，最长的子数组长度为 6。

示例 2：

输入：A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
输出：10
解释：
[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
粗体数字从 0 翻转到 1，最长的子数组长度为 10。

提示：

1 <= A.length <= 20000
0 <= K <= A.length
A[i] 为 0 或 1

通过次数29,484提交次数49

import java.util.Stack;

public class Solution1004 {
	public int longestOnes(int[] A, int K) {
		int max = 0;
		int temp = 0;
		String strA = "";
		String strSub = "";
		String strSubRepK = "";
		String strSubRepALL = "";
		for (int i = 0; i < A.length; i++) {
			strA = strA + String.valueOf(A[i]);
		}
		//System.out.println(strA);
		for (int i = 0; i < strA.length(); i++) {
			for (int j = i + 1; j < strA.length()+1; j++) {
				strSub = strA.substring(i, j);
				//System.out.println(strSub);
				strSubRepK = strSub;
				for (int l = 0; l < K; l++) {
					strSubRepK = strSubRepK.replaceFirst("0", "1");
				}
				strSubRepALL = strSub.replaceAll("0", "1");
				if (strSubRepK.equals(strSubRepALL)) {
					temp = strSub.length();
					if (temp > max) {
						max = temp;
					}

				}
			}

		}
		return max;

	}

	public static void main(String[] args) {

		Solution1004 s = new Solution1004();

		//int[] A = {0, 0, 0, 1 };
		//int K = 4;
		
		
		int[] A = {1,1,1,0,0,1,0,1,0,1,0,1,1,0,0,1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,0,1,1,1,1,1,0,1,1,0,1,1,0,0,0,1,1,0,1,1,1,1,1,1,0,1,0,0,0,0,1,0,1,1,0,1,0,1,0,0,1,1,0,1,0,1,0,1,1,1,0,0,1,0,1,1,1,1,1,1,1,1,1,0,0,0,1,1,0,0,1,0,0,1,1,1,1,1,1,1,1,0,0,0,1,1,1,0,0,0,1,1,0,1,1,1,1,0,1,0,1,0,1,0,1,1,0,0,1,1,1,1,0,1,0,0,0,1,1,0,0,1,0,1,0,1,1,1,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,0,1,1,1,0,1,0,0,1,0,0,0,0,1,1,0,1,1,1,0,0,1,1,0,0,1,0,0,1,0,0,0,0,1,0,0,1,1,0,0,1,1,1,1,0,0,1,0,0,0,0,1,1,1,1,0,0,1,0,0,1,0,0,0,0,0,1,1,0,1,1,0,1,0,0,1,1,1,0,1,0,0,0,0,1,0,0,1,1,0,0,1,0,1,0,1,0,1,1,1,0,0,0,0,1,0,0,1,0,1,1,1,1,1,0,0,1,1,0,1,1,1,0,1,0,0,0,0,1,1,1,1,1,0,0,1,1,1,1,1,0,0,0,0,0,1,1,0,1,0,1,1,0,1,0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,1,0,1,1,1,1,0,0,1,0,1,1,1,0,1,1,0,1,0,0,1,0,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,0,1,0,1,0,1,1,0,0,1,0,0,1,0,1,1,1,1,1,0,1,0,0,0,0,1,1,0,1,1,0,1,1,0,1,0,1,1,1,0,0,1,0,1,1,0,0,0,0,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,1,1,1,0,0,0,1,0,1,1,1,0,0,0,0,0,1,1,0,1,0,0,0,0,1,1,0,0,1,0,1,0,0,1,1,1,0,1,0,1,1,0,0,1,1,0,1,1,0,0,0,0,0,0,0,1,0,1,1,0,1,1,0,0,0,0,0,0,1,0,1,1,0,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,1,0,1,1,1,0,0,1,0,1,1,1,1,0,0,0,0,0,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,0,1,0,0,1,1,1,0,0,0,0,1,0,1,0,0,0,1,0,0,1,1,1,1,1,0,1,0,1,0,0,0,0,0,1,0,0,1,0,0,1,1,0,0,1,1};
				int K = 144;

		System.out.println(s.longestOnes(A, K));
	}
}

中英文超过，感觉功能实现了。