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  • 第四章 分治策略 4.1 最大子数组问题(自己想的,不知道是不是减治法)

    原来不是减治法啊,也算一种新方法了。时间复杂度也是O(n)了。应该是O(3n)
    package
    chap04_Divide_And_Conquer; import static org.junit.Assert.*; import java.util.Arrays; import org.junit.Test; /** * 算反导论第四章 4.1 最大子数组 * * @author xiaojintao * */ public class Maximum_Subarray_Problem { /** * 最大子数组类 left为头部索引,right为尾部索引,sum为数组和 * * @author xiaojintao * */ protected static class MaxSubarray { int left; int right; long sum; public MaxSubarray(int left, int right, long sum) { this.left = left; this.right = right; this.sum = sum; } } /** * 类似减治法,更快速的球最大子数组,支持全部都是负数的数组 * * @param n * @return */ static MaxSubarray findMaxSubarrayFaster(int[] n) { long sum = Long.MIN_VALUE; int right = 0; int left = 0; int negativeCounter = 0; long tempSum = Long.MIN_VALUE; long tempSum1 = 0; // 考虑全是负数 for (int i = 0; i < n.length; i++) { if (n[i] < 0) { if (tempSum < n[i]) { tempSum = n[i]; right = i; } negativeCounter += 1; } else if (n[i] >= 0) { left = i; break; } } if (negativeCounter == n.length) { left = right; return new MaxSubarray(left, right, tempSum); } else { for (int j = left; j < n.length; j++) { tempSum1 += n[j]; if (tempSum1 > sum) { sum = tempSum1; right = j; } } sum = Long.MIN_VALUE; tempSum1 = 0; int tempLeft = left; for (int k = right; k >= tempLeft; k--) { tempSum1 += n[k]; if (tempSum1 > sum) { sum = tempSum1; left = k; } } return new MaxSubarray(left, right, sum); } } /** * 更快的打印最大子数组 * * @param n */ static void printMaxSubarrayFaster(int[] n) { MaxSubarray maxSubarray = findMaxSubarrayFaster(n); int[] result = Arrays.copyOfRange(n, maxSubarray.left, maxSubarray.right + 1); System.out.println("MaxSubarray is " + Arrays.toString(result) + "from" + maxSubarray.left + " to " + maxSubarray.right + ", Sum is " + maxSubarray.sum); } }
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  • 原文地址:https://www.cnblogs.com/xiaojintao/p/3773717.html
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