Leetcode:Largest Number详细题解

题目

Given a list of non negative integers, arrange them such that they form the largest number.

For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330.

Note: The result may be very large, so you need to return a string instead of an integer.

原题链接: https://oj.leetcode.com/problems/largest-number/

算法分析

case1(一般情况):

[3, 30, 34, 5, 9] -> (9 -> 5 -> 34 -> 3 -> 30) -> 9534330

直观想法按位从高到底排序

可以很容易得到9->5的顺序,然而接下来问题来了,位相等的情况怎么办?

考虑3,30,34(数字组1)

简单考虑[3, 30],显然3->30要比30->3的值更大,即3>30的个位0;

再考虑[3, 34],(34->3) > (3->34),即34的个位4>3;

最后[30, 34],34 > 30;

所以数字组1的排序为34->3->30;

最终结果为9->5->34->3->30

case2(不止一位相等,多位高位相等的情况):

[824, 8247] -> (824 -> 8247) -> 8248247

逐一从高位到低位比较,那么第二个数字的最低位7应该与第一个数字的哪位比较呢?决定这两数顺序的不外乎,824->8247,8247->824这两种情况,直观上7应与第一个数字的第一位8比较,由于7<8,所以824->8247

case3 (不止一位相等,多位高位相等的情况):

[824, 82483] -> (82483 -> 824) -> 82483824

case4(重复数字):

[33, 333] -> 33333

一般考虑假设待比较的数字为a1a2, b1b2b3,a1b1…均为位;在重复数字的情况下

a1 a2

||   ||

b1 b2 b3

且b3 == a1,b1 == a2,此时可以得到b1 == a1 == a2 == b2 == b3,即全等,因此最大的比较次数为数字1的位数加数字2的位数 - 1次,该例子的情况为4次。

题目陷阱

case1(有数字为0):

[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

case2(数字均为0):

[0, 0]

算法设计

Integer类,将int按位存储,next取出下一位方法;

 class Integer {
public:
Integer(int i); int getCount() { return count; } int next() {
if (!tmp_count) {
tmp_count = count;
}
return digits[--tmp_count];
} private:
int _i;
int count;
int tmp_count;
int digits[];
}; Integer::Integer(int i):count(),tmp_count() {
// there has a great trap when i == 0
if (i) {
while (i) {
digits[count++] = i % ;
i /= ;
}
} else {
++count;
digits[] = ;
}
tmp_count = count;
}

比较函数cmp,按位从高到低循环比较,等于最大比较次数后退出;

 bool cmp(const int& a, const int& b) {
Integer ia(a);
Integer ib(b); int maxCmpCount = ia.getCount() + ib.getCount() - ;
int curCmpCount = ; while (curCmpCount < maxCmpCount) {
int bita = ia.next();
int bitb = ib.next(); if (bita > bitb) {
return true;
} if (bita < bitb) {
return false;
} ++curCmpCount;
} return false;
}

完整代码(Runtime:9ms)

 #include <string>
#include <vector>
#include <cstdio> class Integer {
public:
Integer(int i); int getCount() { return count; } int next() {
if (!tmp_count) {
tmp_count = count;
}
return digits[--tmp_count];
} private:
int _i;
int count;
int tmp_count;
int digits[];
}; Integer::Integer(int i):count(),tmp_count() { // there has a great trap when i == 0
if (i) {
while (i) {
digits[count++] = i % ;
i /= ;
}
} else {
++count;
digits[] = ;
}
tmp_count = count;
} bool cmp(const int& a, const int& b) {
Integer ia(a);
Integer ib(b); int maxCmpCount = ia.getCount() + ib.getCount() - ;
int curCmpCount = ; while (curCmpCount < maxCmpCount) {
int bita = ia.next();
int bitb = ib.next(); if (bita > bitb) {
return true;
} if (bita < bitb) {
return false;
} ++curCmpCount;
} return false;
} class Solution {
public:
std::string largestNumber(std::vector<int> &num) {
// there is a trap when nums is all zero
bool allZero = true;
for (auto itr = num.begin(); allZero && itr != num.end(); ++itr) {
if (*itr != ) {
allZero = false;
}
} if (allZero) {
return std::string("");
} std::sort(num.begin(), num.end(), cmp);
std::string rel;
char tmp[];
for (auto itr = num.begin(); itr != num.end(); ++itr) {
sprintf(tmp, "%d", *itr);
rel += tmp;
}
return rel;
}
};
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