劍指Offer之Java算法習題精講數(shù)組查找與字符串交集

更新時間：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，反復去看，做過之后可能會忘記，之后再做一次，記不住就反復做，反復尋求思路和規(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字符串所涵蓋的另一個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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