//求最大子数组和,暴力破解法,时间复杂度:O (n ^ 3)
int maxSubArray (int *, int n);
int maxSubArray (int *, int n) {
,,,int 我;
,,,int j;
,,,int k;
,,,int ans =, -100000000;
,,,(小姐:=,0;,小姐:& lt;, n;,我+ +){
,,,,,,,,(=j 我;,j & lt;, n;, j + +) {
,,,,,,,,,,,int sum =, 0;
,,,,,,,,,,,,(=k 我;,k & lt;=, j;, k + +) {
,,,,,,,,,,,,,,,sum +=, [k];
,,,,,,,,,,,}
,,,,,,,,,,,如果(sum 祝辞,ans) {
,,,,,,,,,,,,,,,ans =,总和;
,,,,,,,,,,,}
,,,,,,,}
,,,}
,,,return 答;
}
void 主要(void) {
,,,int 一个[],=,{1,,2,,3,3,5,6,1};
,,,int count =, sizeof (a)/sizeof (int);
,,,int maxNumber;
,,,maxNumber =, maxSubArray(,,数);
,,,printf (" % d \ n ",, maxNumber);
} 结果截图
# include//最大子数字和:贪心算法,时间复杂度为:O (n)
int maxSubArray (int *, int n);
int maxSubArray (int *, int n) {
,,,int 我;
,,,int ans =, -10000000;
,,,int sum =, 0;
,,,(小姐:=,0;,小姐:& lt;, n;,我+ +){
,,,,,,,sum +=,(我);
,,,,,,,如果(sum 祝辞,ans) {
,,,,,,,,,,,ans =,总和;,,//保存先前的最大值
,,,,,,,}
,,,,,,,如果(sum & lt;, 0) {
,,,,,,,,,,,sum =, 0;,//将一部分和& lt; 0的直接删去
,,,,,,,}
,,,}
,,,return 答;
}
void 主要(void) {
,,,int 一个[],=,{1,,2,,3,,,,,,3,,3,,2,,3};
,,,int count =, sizeof (a)/sizeof (int);
,,,int maxNumber;
,,,maxNumber =, maxSubArray(,,数);
,,,printf (" % d \ n ",, maxNumber);
} 结果截图
最大子数组和
