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C語言雙指針——經典題型
2022-07-06 08:32:00 【終為—NULL】
每天進步一點點,堅持帶來大改變!!!
1.序列中删除指定數字
牛客網鏈接:
描述
有一個整數序列(可能有重複的整數),現删除指定的某一個整數,輸出删除指定數字之後的序列,序列中未被删除數字的前後比特置沒有發生改變。
數據範圍:序列長度和序列中的值都滿足 1≤n≤50
輸入描述:
第一行輸入一個整數(0≤N≤50)。
第二行輸入N個整數,輸入用空格分隔的N個整數。
第三行輸入想要進行删除的一個整數。
輸出描述:
輸出為一行,删除指定數字之後的序列。
示例1
輸入:
6
1 2 3 4 5 9
4
輸出:1 2 3 5 9示例2
輸入:
5
1 2 3 4 6
5
輸出:1 2 3 4 6
思路:
定義兩個變量,都從數組下標為0的比特置開始,i變量遍曆整個數組,j變量用來存放不是被删除的元素,當i找到删除的元素之後i繼續向後訪問,j不加加,當不是要删除的元素的時候,將下標為i的元素存放到j下標,然後j繼續加加。
#include<stdio.h> int main() { int n = 0; int arr[50] = { 0 }; scanf("%d", &n); int i = 0; for (i = 0; i < n; i++) { scanf("%d", &arr[i]); } int del = 0; int j = 0; scanf("%d", &del); for (i = 0; i < n; i++) { if (arr[i] != del) { arr[j++] = arr[i]; } } for (i = 0; i < j; i++) { printf("%d ", arr[i]); } return 0; }
2.序列中删除去重
牛客網鏈接:
描述
輸入n個整數的序列,要求對這個序列進行去重操作。所謂去重,是指對這個序列中每個重複出現的整數,只保留該數第一次出現的比特置,删除其餘比特置。
輸入描述:
輸入包含兩行,第一行包含一個正整數n(1 ≤ n ≤ 1000),錶示第二行序列中數字的個數;第二行包含n個整數(範圍1~5000),用空格分隔。
輸出描述:
輸出為一行,按照輸入的順序輸出去重之後的數字,用空格分隔。
示例1
輸入:
5
10 12 93 12 75
輸出:
10 12 93 75
思路:
#include<stdio.h> int main() { int n = 0; scanf("%d", &n); int arr[1000] = { 0 }; int i = 0; for (i = 0; i < n; i++) { scanf("%d", &arr[i]); } for (i = 0; i < n; i++) { int j = 0; for (j = i + 1; j < n; j++) { if (arr[i] == arr[j]) { int k = 0; for (k = j; k < n - 1; k++) { arr[k] = arr[k + 1]; } n--; j--; } } } for (i = 0; i < n; i++) { printf("%d ", arr[i]); } return 0; }
3.有序序列的合並:
牛客網鏈接:
描述
輸入兩個昇序排列的序列,將兩個序列合並為一個有序序列並輸出。
數據範圍:1≤n,m≤1000 , 序列中的值滿足 :0≤val≤30000
輸入描述:
輸入包含三行,
第一行包含兩個正整數n, m,用空格分隔。n錶示第二行第一個昇序序列中數字的個數,m錶示第三行第二個昇序序列中數字的個數。
第二行包含n個整數,用空格分隔。
第三行包含m個整數,用空格分隔。輸出描述:
輸出為一行,輸出長度為n+m的昇序序列,即長度為n的昇序序列和長度為m的昇序序列中的元素重新進行昇序序列排列合並。
示例1
輸入:
5 6
1 3 7 9 22
2 8 10 17 33 44
輸出:1 2 3 7 8 9 10 17 22 33 44
思路:
#include<stdio.h> int main() { int arr1[1000] = { 0 }; int arr2[1000] = { 0 }; int arr3[2000] = { 0 }; int n = 0; int m = 0; scanf("%d%d", &n, &m); int i = 0; for (i = 0; i < n; i++) { scanf("%d", &arr1[i]); } for (i = 0; i < m; i++) { scanf("%d", &arr2[i]); } i = 0; int j = 0; int k = 0; while (i < n && j < m) { if (arr1[i] < arr2[j]) { arr3[k++] = arr1[i++]; } else { arr3[k++] = arr2[j++]; } } if (i == n) { for (; j < m; j++) { arr3[k++] = arr2[j]; } } else { for (; i < n; i++) { arr3[k++] = arr1[i]; } } for (i = 0; i < n + m; i++) { printf("%d ", arr3[i]); } return 0; }
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