劍指Offer之Java算法習(xí)題精講數(shù)組查找與字符串交集

更新時(shí)間：2022年03月22日 10:36:49 作者：明天一定.

跟著思路走，之后從簡單題入手，反復(fù)去看，做過之后可能會忘記，之后再做一次，記不住就反復(fù)做，反復(fù)尋求思路和規(guī)律，慢慢積累就會發(fā)現(xiàn)質(zhì)的變化

題目一

數(shù)組題——二分查找法

寫一個(gè)函數(shù)查找給定的數(shù)組中指定的數(shù)值

具體題目如下

?解法

class Solution {
    public int search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        while(left<=right){
            int mid = left+(right-left)/2;
            if(nums[mid]==target){
                return mid;
            }else if(nums[mid]>target){
                right = mid - 1;
            }else if(nums[mid]<target){
                left = mid + 1;
            }
        }
        return -1;
    }
}

題目二

數(shù)組題——查找數(shù)組中元素位置

根據(jù)給定的數(shù)組按照指定條件查找首尾元素位置

具體題目如下

?解法：

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1,-1};
        int left = 0;
        int right = nums.length-1;
        while(left<=right){
            int mid = left+(right-left)/2;
            if(nums[mid] == target){
                res[0] = mid;
                right = mid-1;
            }else if(nums[mid]>target){
                right = mid - 1;
            }else if(nums[mid]<target){
                left = mid+1;
            }
        }
        left = 0;
        right =  nums.length-1;
        while(left<=right){
            int mid = left+(right-left)/2;
            if(nums[mid] == target){
                res[1] = mid;
                left = mid+1;
            }else if(nums[mid]>target){
                right = mid - 1;
            }else if(nums[mid]<target){
                left = mid+1;
            }
        }
        return res;        
    }
}

題目三

字符串題——查找字符串交集

根據(jù)給定的字符串按照指定條件查找一個(gè)字符串所涵蓋的另一個(gè)字符串字符的最小子串

具體題目如下

解法

class Solution {
    public String minWindow(String s, String t) {
        if(s == null || s == "" || t == null || t == "" || s.length() < t.length()){
            return "";
        }
        int[] need = new int[128];
        int[] have = new int[128];
        for (int i = 0; i < t.length(); i++) {
            need[t.charAt(i)]++;
        }      
        int left = 0, right = 0;
        int  min = s.length() + 1, count = 0, start = 0;
        while (right < s.length()){
            char r = s.charAt(right);
            if (need[r] == 0) {
                right++;
                continue;
            }
            if (have[r] < need[r]) {
                count++;
            }
            have[r]++;
            right++;
            while (count == t.length()){
                if (right - left < min) {
                    min = right - left;
                    start = left;
                }
                char l = s.charAt(left);
                 if (need[l] == 0) {
                    left++;
                    continue;
                }
                 if (have[l] == need[l]) {
                    count--;
                }
                have[l]--;
                left++;
            }
        }
        if (min == s.length() + 1) {
            return "";
        }
        return s.substring(start, start + min);  
    }
}

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