76. Minimum Window Substring

Description

tags: Hash Table, Two Pointers, String difficulty: Hard算法

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).数组

Example:code

Input: S = "ADOBECODEBANC", T = "ABC"  
Output: "BANC"

Note:blog

If there is no such window in S that covers all characters in T, return the empty string "".
If there is such window, you are guaranteed that there will always be only one unique minimum window in S.ip

Solution

  1. 建立一个包含全部T中字节的窗口
  2. 从左侧减少窗口大小,若是符合条件(包含全部T中字节),记录最小长度与起止位置
  3. 若是不符合条件,从右侧扩大窗口,使之符合条件
  4. 重复二、 3直到 S 最末端
func minWindow(s string, t string) string {
    if len(s) == 0 || len(t) == 0 {
        return ""
    }
    mapT := make([]int, 256)
    for i:=0;i<len(t);i++{
        mapT[int(t[i])]+=1
    }
    posPair := []int{-1,-1}
    counter := len(t)
    minWin := len(s)
    
    start :=0
    
    for index,val := range s{
        mapT[int(val)] -=1
        if mapT[int(val)] >= 0{
            counter--
        }
        
        for ;counter==0;start++ {
            if minWin > index - start {
                posPair[0] = start
                posPair[1] = index
                minWin = index - start
            }
            mapT[int(s[start])]+=1
            if mapT[int(s[start])] > 0 {
                counter++
            }
        }
    }
    if posPair[0] == -1 {
        return ""
    }
    return s[posPair[0]:posPair[1]+1]
}

这是一个滑动窗口的算法题,基本的滑动窗口算法窗口大小是固定的,本题是滑动窗口的一个变体应用。element

基本滑动窗口

能够查看Window Sliding Techniqueleetcode

应用方向:求连续大小为K的子数组的最大最小的值,和,积,xor 等字符串

例子:Given an array of integers of size ‘n’. Our aim is to calculate the maximum sum of ‘k’ consecutive elements in the array.get

变体滑动窗口

基本滑动窗口大小是固定的,而通常滑动窗口的大小是须要咱们本身计算的,本题就是一例,根据网友的总结10-line template that can solve most 'substring' problems, 不少的子字符串类题均可以使用这种方法。string

相关文章
相关标签/搜索