leetcode 回溯法解题

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leetcode 17.电话号码的字母组合

class Solution(object):
    def letterCombinations(self, digits):
        if not digits: return []
        res = []
        d = [‘‘, ‘*‘, ‘abc‘, ‘def‘, ‘ghi‘, ‘jkl‘, ‘mno‘, ‘pqrs‘, ‘tuv‘, ‘wxyz‘]
        def dfs(tmp, index):
            if index == len(digits):
                res.append(tmp)
                return
            c = digits[index]
            letters = d[ord(c)-48]
            for i in letters:
                dfs(tmp+i, index+1)
        dfs(‘‘, 0)
        return res
# 求解
s = Solution()
s.letterCombinations(‘23‘)
# return
[‘ad‘, ‘ae‘, ‘af‘, ‘bd‘, ‘be‘, ‘bf‘, ‘cd‘, ‘ce‘, ‘cf‘]

leetcode 22.括号生成

class Solution(object):
    def generateParenthesis(self, n):
        if not n: return []
        res = []
        self.dfs(res, n, n, ‘‘)
        return res   
    def dfs(self, res, left, right, path):
        if left==0 and right==0:
            res.append(path)
            return 
        if left>0:
            self.dfs(res, left-1, right, path+‘(‘)
        if left<right:
            self.dfs(res, left, right-1, path+‘)‘)

leetcode 39.组合总和

class Solution(object):
    def combinationSum(self, candidates, target):
        def backtrack(i, tmp_sum, tmp):
            if tmp_sum>target and i==n: return
            if tmp_sum==target:
                res.append(tmp)
                return
            for j in range(i, n):
                if tmp_sum + candidates[j] > target: break
                backtrack(j, tmp_sum+candidates[j], tmp+[candidates[j]])
        candidates.sort()
        n = len(candidates)
        res = []
        backtrack(0, 0, [])
        return res

leetcode 40.组合总和 II

class Solution(object):
    def combinationSum2(self, candidates, target):
        def backtrack(i, tmp_sum, tmp):
            if tmp_sum>target and i==n: return
            if tmp_sum==target:
                res.append(tmp)
                return
            for j in range(i, n):
                if tmp_sum+candidates[j]>target: break
                if j>i and candidates[j]==candidates[j-1]: continue
                backtrack(j+1, tmp_sum+candidates[j], tmp+[candidates[j]])
        candidates.sort()
        n = len(candidates)
        res = []
        backtrack(0, 0, [])
        return res

leetcode 46.全排列

class Solution(object):
    def permute(self, nums):
        def backtrack(nums, tmp):
            if not nums:
                res.append(tmp)
                return
            for i in range(len(nums)):
                backtrack(nums[:i]+nums[i+1:], tmp+[nums[i]])
        
        res = []
        backtrack(nums, [])
        return res

leetcode 47.全排列 II

class Solution(object):
    def permuteUnique(self, nums):
        def backtrack(nums, tmp):
            if not nums:
                res.append(tmp)
                return
            for i in range(len(nums)):
                if i>0 and nums[i]==nums[i-1]: continue
                backtrack(nums[:i]+nums[i+1:], tmp+[nums[i]])

        nums.sort()
        res = []
        backtrack(nums, [])
        return res

leetcode 51.N皇后

class Solution(object):
    def solveNQueens(self, n):
        # 记录 行, 列, 正对角,负对角,不能有两个以上的棋子.
        # 正对角就是相加之和一样的
        # 负对角就是相减z之差一样的
        def backtrack(i, tmp, col, z_dig, f_dig):
            if i==n:
                res.append(tmp)
                return
            for j in range(n):
                if j not in col and i+j not in z_dig and i-j not in f_dig:
                    backtrack(i+1, tmp+[s[:j]+‘Q‘+s[j+1:]], col | {j}, z_dig | {i+j}, f_dig | {i-j})
        res = []
        s = ‘.‘*n
        backtrack(0, [], set(), set(), set())
        return res

leetcode 52.N皇后 II

class Solution(object):
    def totalNQueens(self, n):
    # 思路与上题类似
        def backtrack(i, col, z_dig, f_dig):
            if i==n: return True
            for j in range(n):
                if j not in col and i+j not in z_dig and i-j not in f_dig:
                    if backtrack(i+1, col|{j}, z_dig|{i+j}, f_dig|{i-j}): self.res += 1
            return False
        self.res = 0
        backtrack(0, set(), set(), set())
        return self.res

leetcode 77.组合

class Solution(object):
    def combine(self, n, k):
        def backtrack(nums, tmp):
            if len(tmp)==k:
                res.append(tmp)
                return
            for i in range(len(nums)):
                backtrack(nums[i+1:], tmp+[nums[i]])
        res = []
        nums = range(1, n+1)
        backtrack(nums, [])
        return res

leetcode 78.子集

class Solution(object):
    def subsets(self, nums):
        def backtrack(nums, tmp):
            res.append(tmp)
            for i in range(len(nums)):
                backtrack(nums[i+1:], tmp+[nums[i]])
        res = []
        backtrack(nums, [])
        return res

leetcode 90.子集 II

class Solution(object):
    def subsetsWithDup(self, nums):
        def backtrack(nums, tmp):
            res.append(tmp)
            for i in range(len(nums)):
                if i>0 and nums[i]==nums[i-1]: continue
                backtrack(nums[i+1:], tmp+[nums[i]])
        res = []
        nums.sort()
        backtrack(nums, [])
        return res

leetcode 131.分割回文串

class Solution(object):
    def partition(self, s):
        def helper(s, tmp):
            if not s:
                res.append(tmp)
                return
            for i in range(1, len(s)+1):
                if s[:i]==s[:i][::-1]:
                    helper(s[i:], tmp+[s[:i]])
        res = []
        helper(s, [])
        return res

leetcode 132.分割回文串 II

class Solution(object):
    def minCut(self, s):
        if s==s[::-1]: return 0
        res = float("inf")
        for i in range(1, len(s)+1):
            if s[:i]==s[:i][::-1]:
                res = min(self.minCut(s[i:])+1, res)
        return res

leetcode 回溯法解题

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原文地址：https://www.cnblogs.com/yutingting/p/12739215.html