代码随想录day23|39组合总数40组合总数II131分割回文串|01笔记
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· 阅读数 4
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39组合总数
- 代码随想录 (programmercarl.com)
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第一印象
- 本题的元素可以重复,那么返回条件就是是大于等于目标值。依旧是利用path和result存储值。
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讲解观后感
- sum值可以不作为参数传入递归函数中,直接用target -。但是保有sum值可以方便剪枝。
- 剪枝操作可以先通过排序使数组单调递增。再利用当前sum与当前值的和来提前终止for循环.
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解题代码
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func combinationSum(candidates []int, target int) [][]int { ans := make([][]int, 0) if len(candidates)==0 { return ans } path := []int{} sum := 0 var dfs func(candidates []int, idx int, sum int, target int) dfs = func(candidates []int, idx int, sum int, target int) { if sum > target { return } if sum == target { temp := make([]int, len(path)) copy(temp, path) ans = append(ans, temp) return } for i:=idx;i<len(candidates);i++ { sum += candidates[i] path = append(path, candidates[i]) dfs(candidates, i, sum, target) path = path[:len(path)-1] sum -= candidates[i] } } dfs(candidates, 0, sum, target) return ans }
- 剪枝:
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func combinationSum(candidates []int, target int) [][]int { ans := make([][]int, 0) if len(candidates)==0 { return ans } path := []int{} sum := 0 sort.Ints(candidates) //排序操作 var dfs func(candidates []int, idx int, sum int, target int) dfs = func(candidates []int, idx int, sum int, target int) { if sum == target { temp := make([]int, len(path)) copy(temp, path) ans = append(ans, temp) return } for i:=idx;i<len(candidates)&&sum+candidates[i]<=target;i++ { //剪枝操作 sum += candidates[i] path = append(path, candidates[i]) dfs(candidates, i, sum, target) path = path[:len(path)-1] sum -= candidates[i] } } dfs(candidates, 0, sum, target) return ans }
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40组合总数II
- 代码随想录 (programmercarl.com)
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第一印象
- 本题每个数字只能使用一次,那么递归传参就得+1
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讲解观后感
- 这题的去重操作十分巧妙。关于树形结构中树层与树枝的讨论十分形象。
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解题代码
- 使用used数组
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func combinationSum2(candidates []int, target int) [][]int { ans := make([][]int, 0) if len(candidates)==0 { return ans } path := []int{} sum := 0 sort.Ints(candidates) //排序操作 used := make([]bool, len(candidates)) var dfs func(candidates []int, target int,idx int, sum int) dfs = func(candidates []int, target int,idx int, sum int) { // if sum > target { // return // } if sum == target { temp := make([]int, len(path)) copy(temp, path) ans = append(ans, temp) return } for i:=idx;i<len(candidates)&&sum+candidates[i]<=target;i++ { // used[i - 1] == true,说明同一树枝candidates[i - 1]使用过 // used[i - 1] == false,说明同一树层candidates[i - 1]使用过 if i > 0 && candidates[i] == candidates[i-1] && used[i-1] == false { continue } sum += candidates[i] path = append(path, candidates[i]) used[i] = true dfs(candidates, target, i+1, sum) used[i] = false path = path[:len(path)-1] sum -= candidates[i] } } dfs(candidates, target, 0, sum) return ans }
- 不使用used数组
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func combinationSum2(candidates []int, target int) [][]int { ans := make([][]int, 0) if len(candidates)==0 { return ans } path := []int{} sum := 0 sort.Ints(candidates) //排序操作 var dfs func(candidates []int, target int,idx int, sum int) dfs = func(candidates []int, target int,idx int, sum int) { if sum == target { temp := make([]int, len(path)) copy(temp, path) ans = append(ans, temp) return } for i:=idx;i<len(candidates)&&sum+candidates[i]<=target;i++ { // i != start 限制了这不对深度遍历到达的此值去重 if i != idx && candidates[i] == candidates[i-1] { // 去重 continue } sum += candidates[i] path = append(path, candidates[i]) dfs(candidates, target, i+1, sum) path = path[:len(path)-1] sum -= candidates[i] } } dfs(candidates, target, 0, sum) return ans }
- 不使用sum传参
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var ( res [][]int path []int ) func combinationSum2(candidates []int, target int) [][]int { res, path = make([][]int, 0), make([]int, 0, len(candidates)) sort.Ints(candidates) // 排序,为剪枝做准备 dfs(candidates, 0, target) return res } func dfs(candidates []int, start int, target int) { if target == 0 { // target 不断减小,如果为0说明达到了目标值 tmp := make([]int, len(path)) copy(tmp, path) res = append(res, tmp) return } for i := start; i < len(candidates); i++ { if candidates[i] > target { // 剪枝,提前返回 break } // i != start 限制了这不对深度遍历到达的此值去重 if i != start && candidates[i] == candidates[i-1] { // 去重 continue } path = append(path, candidates[i]) dfs(candidates, i+1, target - candidates[i]) path = path[:len(path) - 1] }
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131分割回文串
- 代码随想录 (programmercarl.com)
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讲解观后感
- 本题也是经典的利用回溯算法的题目。而关于回溯的用法中,首先遇到的难点就是分割究竟应该如何表示。只要把index和i构成的区间是分割后的区间理解明白,就能解决了。
- 卡尔也帮我们列出了这道题的几个难点,帮助我们理解。
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切割问题可以抽象为组合问题 如何模拟那些切割线 切割问题中递归如何终止 在递归循环中如何截取子串 如何判断回文
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解题代码
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var ( path []string // 放已经回文的子串 res [][]string ) func partition(s string) [][]string { path, res = make([]string, 0), make([][]string, 0) dfs(s, 0) return res } func dfs(s string, start int) { if start == len(s) { // 如果起始位置等于s的大小,说明已经找到了一组分割方案了 tmp := make([]string, len(path)) copy(tmp, path) res = append(res, tmp) return } for i := start; i < len(s); i++ { str := s[start : i+1] if isPalindrome(str) { // 是回文子串 path = append(path, str) dfs(s, i+1) // 寻找i+1为起始位置的子串 path = path[:len(path)-1] // 回溯过程,弹出本次已经填在的子串 } } } func isPalindrome(s string) bool { for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 { if s[i] != s[j] { return false } } return true }