具有1000位数的字符串，使用substring方法查找没有数组的最大5位数字

Question

我知道我的代码可以变得更简单，更有效但是我试图让它以这种方式工作。

我希望得到字符串中包含5位数字的最大数字。我想使用public class thousandDigits { public static void main(String[] args) { String num = new String ("73167176531330624919225119674426574742355349194934" + "96983520312774506326239578318016984801869478851843" + "85861560789112949495459501737958331952853208805511" + "12540698747158523863050715693290963295227443043557" + "66896648950445244523161731856403098711121722383113" + "62229893423380308135336276614282806444486645238749" + "30358907296290491560440772390713810515859307960866" + "70172427121883998797908792274921901699720888093776" + "65727333001053367881220235421809751254540594752243" + "52584907711670556013604839586446706324415722155397" + "53697817977846174064955149290862569321978468622482" + "83972241375657056057490261407972968652414535100474" + "82166370484403199890008895243450658541227588666881" + "16427171479924442928230863465674813919123162824586" + "17866458359124566529476545682848912883142607690042" + "24219022671055626321111109370544217506941658960408" + "07198403850962455444362981230987879927244284909188" + "84580156166097919133875499200524063689912560717606" + "05886116467109405077541002256983155200055935729725" + "71636269561882670428252483600823257530420752963450") ; int greatest = 0; int max = -1; int numChar = Integer.parseInt(num.substring(0, 5)); for (int i = 0; i < num.length() - 5; i++) { greatest = numChar; if (max < greatest) { max = greatest; } } System.out.print(max); } } 方法。作为输出，我只得到前5个数字，并且它不会遍历字符串的其余部分。

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<usertask xmlns="..."></usertask><taskid></taskid><blablabla></blablabla></usertask>

输出为7316但是应该是99890作为最大的5位数字。

Answer 1

那是因为你没有更新循环内的numChar。你只在开始时做一次。此外，计算出的substring需要在循环中从i转到i+5。

int greatest = 0;
int max = -1;
//int numChar = Integer.parseInt(num.substring(0, 5)); <-- Not Here

for (int i = 0; i < num.length() - 5; i++) {
    int numChar = Integer.parseInt(num.substring(i, i + 5)); //  <-- but here
    greatest = numChar;

    if (max < greatest) {
        max = greatest;
    }
}

Answer 2

您不需要解析该值，因为字符串比较没问题。

你可以做到

String num = "73167176531330624919225119674426574742355349194934" +
        "96983520312774506326239578318016984801869478851843" +
        "85861560789112949495459501737958331952853208805511" +
        "12540698747158523863050715693290963295227443043557" +
        "66896648950445244523161731856403098711121722383113" +
        "62229893423380308135336276614282806444486645238749" +
        "30358907296290491560440772390713810515859307960866" +
        "70172427121883998797908792274921901699720888093776" +
        "65727333001053367881220235421809751254540594752243" +
        "52584907711670556013604839586446706324415722155397" +
        "53697817977846174064955149290862569321978468622482" +
        "83972241375657056057490261407972968652414535100474" +
        "82166370484403199890008895243450658541227588666881" +
        "16427171479924442928230863465674813919123162824586" +
        "17866458359124566529476545682848912883142607690042" +
        "24219022671055626321111109370544217506941658960408" +
        "07198403850962455444362981230987879927244284909188" +
        "84580156166097919133875499200524063689912560717606" +
        "05886116467109405077541002256983155200055935729725" +
        "71636269561882670428252483600823257530420752963450";
String largest = IntStream.range(0, num.length() - 5)
        .mapToObj(i -> num.substring(i, i + 5))
        .max(Comparator.<String>naturalOrder())
        .orElseThrow(AssertionError::new);
System.out.println(largest);

打印

 99890