[LeetCode]Summary Ranges

[LeetCode]Summary Ranges 

题目描述：

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].

题目大意：

给定一组排好序且无重复的整数，返回整数范围的汇总。

例如给定数组 [0,1,2,4,5,7]， 返回 ["0->2","4->5","7"]。

解题思路：

将逐一递增的整数序列化简为（起点->终点）即可。

Python代码：

class Solution:
    # @param {integer[]} nums
    # @return {string[]}
    def summaryRanges(self, nums):
        x, size = 0, len(nums)
        ans = []
        while x < size:
            c, r = x, str(nums[x])
            while x + 1 < size and nums[x + 1] - nums[x] == 1:
                x += 1
            if x > c:
                r += "->" + str(nums[x])
            ans.append(r)
            x += 1
        return ans
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下面的Python代码短小精悍，摘自LeetCode Discuss

链接地址：https://leetcode.com/discuss/42199/6-lines-in-python

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Solution 1

Just collect the ranges, then format and return them.

def summaryRanges(self, nums):
    ranges = []
    for n in nums:
        if not ranges or n > ranges[-1][-1] + 1:
            ranges += [],
        ranges[-1][1:] = n,
    return ['->'.join(map(str, r)) for r in ranges]
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Solution 2

A variation of solution 1, holding the current range in an extra variable r to make things easier.

def summaryRanges(self, nums):
    ranges, r = [], []
    for n in nums:
        if n-1 not in r:
            r = []
            ranges += r,
        r[1:] = n,
    return ['->'.join(map(str, r)) for r in ranges]
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Solution 3

A tricky short version.

def summaryRanges(self, nums):
    ranges = r = []
    for n in nums:
        if `n-1` not in r:
            r = []
            ranges += r,
        r[1:] = `n`,
    return map('->'.join, ranges)
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本文链接：http://bookshadow.com/weblog/2015/06/26/leetcode-summary-ranges/
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